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[Music] what is an epsilon delta argument and why do we need this

why do we need a precise definition of a limit

when i can just use a table of values or a graph to find a limit

when you want to find a limit and have no doubt about it

or when you want to write a proof of a limit theorem

you can use an epsilon delta argument in this video i discuss the limit

definition the precise definition of a limit and show you

how to use it welcome back everyone i’m dave

and in this video we’re going to talk all about the limit definition

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so i’m going to begin by motivating why we need it

then i’m going to give the a precise exact definition of a limit

then we’re going to work through lots of examples on using this limit definition

and then we’re going to prove some theorems using the limit definition

and then stick around to the end where we have some exercises over

the limit definition so let’s get started okay so up first is why do we need a

rigorous definition uh let’s talk about that

um now when we talk about this though keep in mind though that the first

episode in this series calculus one explore discover learn series um

is a video over the intuitive meaning of the limit and so uh find some limits um

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finding limits now a link to the playlist is in the description below so

you can find that episode there but this is certainly not the first

episode that you should see about limits

so we should have some type of intuitive idea of what a limit is

and this is going to be the precise definition of a limit

so when we’re studying limits we often need a three-pronged approach

so often looking numerically by cons constructing tables of values

so this is especially nice if your function is especially difficult to work with

or you don’t have a good handle on it [Music]

and then a graphical approach if you have some kind of idea of

what the graph looks like or you have some technology at your disposal

finding a limit that way could be a very useful

of course an analytic approach using some algebra

or some limit theorems is is another approach

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um so the precise definition of a limit is is i’m about to talk about and i’m

going to work through a lot uh and first i’m going to talk about some examples

about why the definition is needed so and and now you’ve already made seen

examples of estimating limits numerically and graphically so

if you haven’t check out that video on that

the first episode of the series so each of these approaches

produces an estimation of the limit it gives an estimate of the limit

so the precise definition of a limit gives you the exact value that the limit is

assuming the limit exists of course and also it allows us to prove results

like limit theorems so that you can use analytic techniques

or evaluating limits and those analytic techniques allow you to

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find a lot more limits and a lot faster and so the the uh precise definition is

used a great deal in actually trying to prove theorems

and not just limit theorems but actually all kinds of limit uh all kinds of

theorems and real analysis which is the subject that

calculus belongs to in any case let’s um look at a

an example which kind of motivates why the

definition is needed so let’s look at a an example real quick

so consider us trying to find the value of this limit here

so this is we’re approaching zero and if you notice the numerator is

um you know when you plug in x 0 into the numerator you’re going to get 2

minus 0 minus 2 you’re going to get 0 over 0

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but you know that makes the limit harder to or not harder but it just makes it

much more fun to actually find the limit but i’m going

to look at a table of values and so for my x’s i’m going to choose

negative 0.5 negative 0.1 negative 0.001 and then

i choose some values that are positive 0.001 and 0.005

so those are my choices for the x now how do i get those choices

well in order to find the limit you need to have infinitely many x

values in the domain of your function not including zero itself necessarily

um so i chose x is zero there in my table but i’m not going to be able to get an

output for that x but the values to the left of zero there

are for computing the limit on the left side and the values

on the right for the right side now for each of those

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x values that i choose i substitute them into the function

so i i substitute in negative 0.5 and i substitute that value in so just

to make sure that i don’t lose anybody here i’m going to

take my function here which is f of x here

which is 2 square root of x plus 1 minus x minus 2

over x squared that’s my function f that i’m using for f

and now if i substitute in my first x value that i chose negative 0.5

i have to substitute that in so i have 2 square root of negative 0.5

plus 1 minus the input 0.5 minus 2 all over negative 0.5 squared

so i’m substituting this x value in to the function

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and i grind away and calculate that and i get out negative 0.3431

so i do all that and i put the value in the table right here in the first

column and then i go repeat that process and the next x value i choose is

negative 0.1 and when i plug in negative 0.1 into the function i get out

negative 0.2633 all right so i choose some x values to the left of 0

and those x values are getting closer and closer to zero

and i choose some values to the right of

zero and those values are getting closer and closer to zero

so at zero of course we’re undefined there is no output there

okay so the number l is suggested to be if we look at the

remember we’re choosing the x values that are getting closer and closer to zero

and we’re asking the question what are the outputs approaching

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and if you look at the outputs right we’re we’re getting closer and closer to

negative 0.25 so you know we have negative 0.2501

and we have negative 0.2499 so it looks like we’re

looks like a limit l is negative 0.25 now i chose some x values here but i

didn’t choose them actually that close to zero i mean come on negative 0.001

isn’t that close to zero if i try something even closer to zero

you might think i might get a different estimation about what the limit might be

so what if we try zero point zero zero zero

zero just to make sure i’ve taken enough numbers

close enough to zero there’s a problem with that though

so you may find the calculator gives 0. so what do i mean by that so

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if i plug in f of let’s see here one we’re going one two three four five

six so one two three four five six one so now i’m going to do

two times square root so this is really really close to zero here

so you may not be convinced that we did enough numbers close to zero that’s not

that close to zero this is much closer to zero and we’re trying to be very clear

about what the limit is so one two three four five six one

so that’s our x x plus one so we have all that plus one now minus the x

so minus this right here point one two three four five six one

and then a minus two you can see the function right there

minus two so there’s all of that over the x squared so zero point one two three

four five six one and then square it so the point of all this is the

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following is that whatever computing device or hopefully you’re not

doing that by hand but whatever whatever device you’re using

to calculate this output here may give you the wrong answer it may it

may just say i don’t have enough decimal accuracy

to give you anything other than just zero for this

so it may just calculate zero for the numerator even though it’s not zero

so if you’re going gonna get zero for the

for the top and then the calculator may give you

who knows what for the denominator so your calculator actually may tell you

zero for this even though if you plug in this number

right here and you calculate it that number is not zero the calculator

may tell you it is exactly zero so if you get really really close to zero

you might be misled into thinking the output is zero

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and so then you might be misled to thinking that the limit is actually zero

so um is the limit zero well no the calculator may give you a

false answer because when x is small enough

when it’s close enough to zero then that expression right

there seems like zero now i know what you’re thinking

you’re thinking oh just go get a more accurate calculator

this number right here for some calculators

for example try your try your iphone and calculate all that up if you can

or you know whatever mobile phone you have but

if if you if you have a good calculator you may be able to calculate this number

out and you won’t get zero you may be able to calculate it exactly right but

what if you put instead of six zeros right here what if we put 30

then this expression by that computing by some computing devices may come out

to be zero so the solution really isn’t to get more powerful computers

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so whatever technology you’re using it’s still that expression still may

seem like zero of course it’s not zero so the point is that using technology to

verify a a limit can lead to misunderstanding so in fact a formal definition

of a limit is needed so using the formal definition of a limit

we can prove what the value of the limit is the value of l without any doubt

any doubt whatsoever now the proof of uh proving a limit is

true using a limit definition that proof is usually called an epsilon

delta proof or sometimes an epsilon delta argument

since the formal definition of a limit is usually stated with the greek letters

epsilon and delta and that’s the main topic of this video

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is that formal definition and so here’s an example of a limit on

this example right here where using technology might lead to

incorrect results um now i’m going to give you um an argument from two people

who don’t know the formal definition and they’re going to have a discussion

and try to figure out what this limit is not knowing the formal definition

[Music] now when i’m looking at this right here i’m going to say hi i’m epi

so this is epi and this right here is delhi hi i’m delhi

so deli’s this person down here and epi’s this person up here and

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they’re going to have an argument over what they think this limit is

okay so epi in delhi here we go [Music] so first epi says i think l

the limit is 6.3 because 2.1 is close to 2. so epi knows you need to choose

something close to 2 and look at the output value so 3 times 2.1

is 6.3 so epi says i i have the limit i think it is 6.3 then delhi says well

2.01 is closer to 2 so why don’t you think the limit is three times

two point zero one which is six point zero three why don’t you think

so basically delhi’s challenging eppie and saying

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uh why don’t you think the limit is 6.03 delhi says i think the limit is

l is six i think it’s six so delhi did some thinking and she says

in any case i think l is six and then eppy replies back oh i see we need

really close to two so now i think l is six point zero three

because two point zero one is really close to

two okay and then delhi comes back and says well

two point zero one is closer to two but i still think l is six i still think

the limit is six i don’t think it is 6.03 as you say

have you considered that 2.001 is even closer to 2.

and if he says no i haven’t considered that yet

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and then after some thinking if he says i’m not sure we’ll ever come to an

agreement on what this limit is so after

um one back and forth and then delhi you know

says you know have you considered this right and

i’m thinking heavy says maybe you can always tell me something that i haven’t

considered so epi evie thinks about that and says

you know i’m not sure we’ll ever come to an agreement on what this limit is

in fact after even some more thinking hebby says

maybe there’s not even one value this limit could be

so delhi says i still think that l is six because any real number that

you give me more or less than six i can help you

realize that your guess is wrong and then delhi says do you want to see how

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and then without much time to respond delhi says

first you give me your new updated guess say it is 6.003

now i am convinced delhi goes on to say that l equals 6 so i use

the difference between those six minus six point zero zero three

and that’s going to be six zero point zero zero three

now deli says i save this number and i’m going to call it epsilon

and i’m going to use this epsilon to convince you

that your guess was wrong your guess is 6.003

daily’s thinking your guess is 6.003 and it’s 0.003 off of my guess

and i’m going to use how much you’re off by and i’m going to convince you that

your guess is wrong so she goes on to say here’s how

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i will reply back with a real number which

i’m going to call delta that will allow you to be sure that your guess is wrong

so i’m going to convince you that your guess is wrong so in other words

i’m going to use your guess and i’m going to compute this epsilon and i’m

going to use that epsilon to come up with the delta and that delta

is going to prove you wrong so you’ll you’ll understand that your guess is wrong

that’s what she’s thinking okay so epi says

okay i’m willing to try this but then epi thinks

and says okay but i’m going to give you an even better guess

what if i give you ls zero 6.00 three is my guess and then epp is thinking

to himself he says i’m using two point zero zero zero one is close to two

that’s pretty close to two let’s see what delhi can say about that

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right and so so eppy says what if i give you 6.0003 is my guess

and then deli quickly responds with her epsilon deltas she says for this guess

i’m gonna use epsilon to be six minus six point zero zero zero three

and that’s how much your guess is off of mine which i think it is which is six

and i’m going to use and and i tell you to consider this number here delta

0.0003 and she’s going to divide by 30 and she’s going to get 0.00001

and her argument is this she she says this you see 2.00001

is closer to 2 than yours so i think 3 times 2.00001

which is six point zero zero zero zero three is a better guess

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and as del as epi is shaking his head yeah that’s a better guess

delhi quickly responds i can do this every time you guess

every time you guess i can find my epsilon and use this epsilon and divide by 30

and i believe that i can convince you that your guess

was wrong no matter what your guess is i can do this argument because i can do

this argument no matter what you give me no matter how small

of a number that you give me away from six

i can give you this argument so in fact that’s why i believe

that l equals six okay so that’s epi and delhi’s argument

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and we’re going to now look at the precise definition of a limit [Music]

now before we look at the precise definition of a limit

i’m going to quickly show you what you see when you’re looking at the intuitive

version of the limit definition and so you know i want to contrast the

two what was the intuitive meaning of a limit

and what’s the formal meaning of a limit so

concisely a limit is used to describe the behavior of a function near a point

but not at the point the function need not even be defined at a point

so i just want to quickly remember remind you of that so for example if i have a

function through here and someone took away

a number right here point on the graph and put the height here

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this is still a function and this limit here this l the height right here is l

and you know there may there might be a hole here there that’s what this is

saying is that the limits used to describe the behavior as we’re approaching c

from both sides we’re approaching c what is the height approaching

the function need not be defined there because this is describing the behavior

of a function the behavior the function is approaching

l it doesn’t matter if we’re equal to l that’s an issue about continuity which

we’re not talking about in this video this episode this episode is about

the limit only so in terms of the limit only we’re not concerned about

being defined at the point but rather as we’re approaching

the point so this is the intuitive meaning of a limit um the value

of the function at the point does not affect the limit but here’s the

intuitive meaning of a limit the intuitive leaning means we can make f of x

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as close to l as we wish by taking x sufficiently close to but different from c

so in other words as we’re approaching as x’s are approaching c

the f of x are approaching l and we can get as close to l

we as we wish all you got to do is choose x x is closer and closer to c

and you’ll get as close to l as you need to using f of x

all right and so this was the intuitive meaning of a limit and now

the formal definition right here so here is the formal definition and i want

to go through this um piece by piece to make sure that you

understand everything here so first off we need a function defined on an open

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interval so that’s crucial so back to our diagram here um we may have a

hole here or not or it may be defined somewhere else

or may just be empty or in fact it may pass through

none of that really matters for the limit it doesn’t matter if it’s just

nice and smooth and continuous or if there’s a hole or if there’s a

hole and then a point somewhere else it none of that matters for the limit so

i’ll just leave there there’s a hole there but

we need a limit we need a function and it needs to be defined

on an interval containing c so we can put an interval here and just call it i

and the reason why f needs to be defined on this interval here

is that we need to be able to choose a lot of x’s in here close to c

and a lot of x’s in here close to c so you have to have

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that happening you need to have an open interval around

around c in which f of x is defined okay now l is a real number

and we’re going to say this limit is equal to l

means so in other words we’re now defining this notation right here the limit as

x approaches c of f of x equals l we’re defining what all this right here

means in terms of these uh this statement here and the statement here

can also be written like this so the the word right there for all

epsilon greater than zero you can also write it like this for all

so this is looks like an upside down a so it says for all epsilon greater than

greater than zero so if you want to write it in words

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for all epsilon greater than zero and you can

some people write the symbol here in words like we did over here

or some people write it in symbols like this so this means for

all and then this is epsilon and then greater than and then

zero so this is shorthand or mathematical notation for these words

and then we have there exists a delta greater than zero

so let’s look at that right here so there exists a delta greater than zero

so here i wrote this symbol looks looks like a backwards

e and the words are there exists a delta we’re using the greek

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letter uh here delta there existed delta um greater than zero

okay so so far we’ve broken down um those words for all eps on grain zero

there existed delta greater than zero and here’s the symbols for this [Music]

and then we have such that okay those that’s words already

and then now we have um this line here and we have all those symbols in there

so i want to break all those down for you

in case you’ve never seen them before so let’s work up here so first off is this

one um actually um this sentence right here has um have some

value in it so i just want to refresh your memory on absolute value

so this right here is equal to um a minus b

if a minus b is greater than or equal to zero

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or it’s equal to negative a minus b if a minus b

is less than zero so this is just the way absolute value works

the absolute value of something is always positive if it’s already a minus

b is already positive then a the absolute value of a minus b is just a minus b

but if a minus b is negative then i have to put a negative in front

of the thing that’s negative to make sure that a negative times a

negative will be positive so now this is a way to think about

uh the absolute value as a piecewise function but

another way to think about um a minus b with

absolute value on it is you know you have a and b

on the number line and you know you could have b and a or you could have a and b

but the absolute value this is you know geometrically this right here

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is going to represent the distance between the distance

between a and b between a and b so you want to think about this as

the distance between a and b and that’s just intuitive

meaning of this right here whereas this right here is analytic

meaning but if you were to go compute the distance between any two numbers on

the real line this is what you have here so

you know i’m going to use the distance between a and b

now in this statement over here we have an x and a c

and then we also have f of x and an l so when i write it with this right here

think of this as the distance between x and c

so that’s what this right here means the distance between x and c

and then if i have an f of x and an l so the distance

and this is taking place on the y axis so this will be the distance between f

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of x and l that’s how you want to interpret this

is the distance between these two y values here

and so this is just the distance between them

all right so now let’s um go back and think about what the sentence here means

this sentence right here uh first off it says

0 is less than the absolute value of x minus c so

what this right here represents is that x is not so

x is not c x is different from c um so that means the distance between

x and c is greater than zero in other words x and c are not the same numbers

okay so that’s one way you want to interpret this right here

um and then we also have um this one right here is less than delta

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so this is saying the distance between x and c is less than delta so distance

between x and c is less than delta so this

this these symbols right here just means

this in words the distance between x and c is less than delta okay now

i’m i’m going to skip over the arrow symbol for a moment

and look at this one right here we have the absolute value of f of x minus l is

less than epsilon okay and so what this represents is this distance between

f of x and l these two right here the distance

between these two is less than the epsilon epsilon okay

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so um now the arrow symbol now this right here means um implication

and so if you have something like p implies q this means if p then q

and without going into logical details here um it just simply means if and then

and at this point and the calculus students career we’re relying upon their

intuitive meaning of this uh here um now if if you’ve had some

proof writing practice then you should be very familiar with this right here

otherwise we’re not going to give a formal definition of what this means

it just simply means if then and that’s the way

so we can write this all out now in terms of words let’s do that

so here it is in symbols here again let’s go over here and

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do that here so we have the limit as x approaches c of f of x

and so this is notation right here that is very useful when you’re finding

limits however when you’re working with proofs

here’s the meaning of a limit for all epsilon greater than zero

there exists a delta grade zero um such that and sometimes abbreviate

that such that or there’s even a symbol there for that such that

the distance between x and c is positive and less than delta implies so if that

happens then the distance between f of x and l is less than epsilon

now at this point i’m just trying to get you familiar with the symbols

so that we can then get a um a better intuitive meaning of the limit

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in terms of geometry um for these epsilon delta so i’m gonna

i’m gonna talk about that in a second but

i just want to get you familiar with all of these symbols here

and so i’m going to write this all out in words the

the statement here is in part part words part symbols here

so let’s put it all in terms of words now you know what are the words for all

these symbols here now so what do we have here for all epsilon greater than zero

there exists a delta greater than zero such that such that if the distance if

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if the distance between x and c is positive and less than delta then

the distance between f of x and l is less than epsilon

okay so if you wrote all these symbols out here

in other words you basically get a paragraph i i think that’s really

you know trying to help you get familiar with the symbols

and what the words mean but another point is

is that this is very useful for making computations right deriving values of

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limits by doing applying limit theorems which we did in the last

episode so you know sometimes you want to work with your function like

factoring it or applying some trig identities you know you want to do some

computational type work this is very useful notation here it’s very

very brief this is very useful for writing proofs

and proofs are important because they prove theorems

which make computations faster which means you can get a lot more work done

and you can understand things a lot better because you have a precise definition

so you can see exactly when things don’t work and exactly when things do work

now all of it has meaning that you can write out in

just regular everyday words but as you can see that is very cumbersome um and so

you know you want to be aware of all three what does it mean in terms of just

regular words how do you use the symbols what does the symbols mean

00:37

and how do we use this and what does this mean

now so we have notation we have the symbols we have the words

now let’s talk about the geometry of it um so this right here

is the same thing as all this is the same thing as all this it’s basically

the idea here you need to be aware of all of these things here

now in terms of the geometry though we have a geometric um

interpretation of this statement here and

when we are looking at this diagram here we we’re going to read through this

statement here and what we’re saying is that for all

epsilon greater than 0 that’s given to us when you’re thinking about epsilon you

want to think about small numbers like point one or point zero zero one or

00:38

point zero zero zero zero zero one but the point is that you wanna

think about really small numbers but they’re greater than zero

so but epsilon is arbitrary to be clear epsilon

is any number any real number greater than zero

and you want to show that the delta has to exist

such that that implication holds and so when we’re looking at this

diagram over here what we’re looking at here is the fact that this

right here i said was the distance between x and c is less than delta

but we can write that out um without the absolute value as this right here and

if we add a c to both to all three sides we’re getting

00:39

c minus delta and then x and then c plus delta and the same thing

for f of x minus l that is less than epsilon what does that

mean this is a very concise way of writing things

this is a concise way of writing things this is if you want to spell it out

what is the x between between these what is the f of x between here so f of x

minus l is less than epsilon but it’s greater than minus epsilon

and so adding an l through on everything we have this right here so this is the

longer version of it but it’s very useful because when you plug in an

x you have a bound for it when you choose an x you have a bound you have a

bound for it here okay so back to the geometry here

so we have c right here is in the middle of c minus delta and c plus delta

00:40

and the l is right between the l plus epsilon and the l minus epsilon

and so what we’re saying here is that no matter what epsilon is given

i can find a delta which gives me an an interval down here

such that if i choose an x any x that satisfies uh this hypothesis here

then the f of x will have to be in this window here

so if i try to draw that out here so we have some function coming through here

and like i said here we may have a hole here but here’s c

and so let’s read through this these symbols here

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so for all epsilon given so here’s our l so no matter what epsilon is given

there’s l plus epsilon and l minus epsilon

so epsilon can be really really small so so i draw i draw it

i drew it kind of big here but epsilon but i zoomed in a lot right

so though this window right here could be really really small epsilon

could be point zero zero zero zero one so that could be a really small distance

there and and and those distance match right there because that’s

plus epsilon and minus epsilon and an epsilon can be any

any real number greater than zero so that could be a really small distance

right here so i drew it kind of big but you know epsilon could be

zero point and then a million zeros and then a one

so that right there that distance right there can be arbitrarily small

but you give me any arbitrarily small distance there

we will come up with a delta there exists a delta

so now i have c plus a delta and a c minus the delta

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so i’ll come up with a an interval here and any x we choose in here so

you know any any x where the distance between x and c is

less than delta so in other words the x is in here think of the x’s in

here somewhere it could be over here over here but it’s the distance between

x and c is less than the delta because you know this this distance right here

is the delta so the x is in here so the distance between

x and c is less than the delta but x is not c so that has to imply

that the distance between f of x and l is less than the epsilon

so that means the f of x is in here somewhere it could be up here

down here so since epsilon is arbitrary this right here is getting squeezed down

00:43

as arbitrarily small as we want no matter what epsilon is given

no matter how thin no matter how small this interval is here

no matter how small it is because it could be for any epsilon

as long as it’s positive so it could be point zero zero with a billion zeros and

then a one no matter how small that gets if there’s always a

delta where i can choose an x in here that guarantees that my f of x is in

here then that f of x that limit of f of x has to be equal to l

so this is the geometric description here this is equal to l

so you know we need to understand the symbols

we need to understand the notation and we need to understand

the geometry this is our function l and you know this is a process that’s

happening it’s not set in stone and the way you

00:44

want to think about the process is epsilon is arbitrary it can be point

zero zero and then a google plex of zeros and then a one

i mean this epsilon can be as small as you want it to be

but in order for the limit to be equal to l you have to always be

able to produce a delta so that if you choose an

x in here f of x will be within this window here and so if i keep squeezing

this down f of x is getting closer and closer to that l

so that’s the definition there the precise definition is given in symbols

and then we use this notation right here for calculations and things like that

[Music] but ultimately relying upon a good geometric

uh your intuition for the epsilons and deltas that that could also be very

00:45

very critical for some problems okay so now let’s go on to the next part [Music]

examples using the epsilon delta definition [Music]

now anytime you see a new definition you should see some you know motivation

as to maybe why you need the definition and then you should see some examples of

using the definition um of course if you’re undergraduate

now if you’re a graduate student then you’re going to be expected to provide

some of those examples on your own in any case we want to see some examples

now and i’m going to guide you through some examples here

of how using an epsilon definition are using an epsilon delta argument

to prove a limit is is is true the value of a limit is equal to an l

using an epsilon delta argument here and um when we when we work these examples

00:46

out i want to have two steps in mind that we need to write out a proof

now actually i said that wrong it’s a two-step process to prove

it’s not a two-step process to write the proof

so think about this as you’re writing a a story or a poem or

you know any short story you want multiple steps

you’re not going to just take a pencil you know pen and paper out

and write down the perfect poem well maybe you will maybe you won’t but

for most people you need to write a draft first so you need to have a process

especially for a long book but any case when you’re writing something out that’s

well thought out you often need multiple steps

so we’re going to write a proof out but that proof

is actually going to be the last step so step number one so this is steps

00:47

for proving the limit as we approach some c

equals l so this will be the steps for proving this

so notice in the first video i used the word finding a lot

find the limit find the limit and you know in that case you may say oh the limit

doesn’t exist right here so here we’re proving a limit so we know what l

is equal to and we want to prove that that that statement is true

so the first step will be find delta we want to find delta so remember this

right here means for all epsilon green zero there exists

a delta greater than zero such that and then there’s other stuff

so this is how it begins we have to start our proof with an arbitrary epsilon is

given any any epsilon is given so imagine someone giving you

00:48

an epsilon and you have to reply back with a delta

so that’s the first step is how do you do that

how do you do this part right here so that’s step number one you so you

want to say here let epsilon greater zero be given

that’ll be the first step in your proof is let epsilon be given

and then you want to use the you want to use the inequality here

f of x minus l is less than epsilon you want to use this right here inequality

right here to find a condition of the form x minus c is less than delta

so you want to find that delta where you have some condition here

00:49

and the in the and the goal here is where the delta here only depends upon

on epsilon in other words delta cannot depend upon x

that’s a delta okay so we want to start by looking at this

and we know what the f of x is that’s given we know what the l is that’s given

and we want to start with this inequality here and write it out

remember epsilon is given so we know what that is also

epsilon is just epsilon it’s given to us so we know that we know that we know

that we know all of this what we don’t know is the delta so we

want to work with this out and make it look something like this and

whatever that delta is that’ll be our new delta and the delta

may have an epsilon in it because we’re starting with this right

here and there’s epsilon in this right here now

00:50

once you find the delta then the second step will be to actually write a proof

write a proof so you can think of this step one here

doing all this right here as scratch work it’s not part of our

final solution that we’re going to write up it’s not part of our final proof

so for example you may write a story about a hero and then in one of your

drafts the hero dies early but then in your final work

the hero doesn’t die to the end well you don’t want to make your hero die twice

it probably won’t make sense so this is the scratch work and you don’t

want to include that when you write your final proof this is just to help you

write a final proof this is the brainstorming this is the

draft writing whatever you want to call it that part right there is to help you

write a proof so now we want to write a proof

00:51

and here’s how your proof should look this just like a template

so you want to start off with for any epsilon greater than zero [Music]

assume that this is true remember because we have an if then so

this is our hypothesis here and we’re knowing what the delta is

because we found it in step one so we know what the delta is and so now

we can say assume this is true and use the relationship between the

epsilon and delta to prove that the distance between f of x and l

is less than epsilon so your proof should start with an epsilon and

end with an epsilon and somewhere in between

00:52

you have to use this delta to go from delta to this to this so you so

you want to start with epsilon’s given and start with this for the delta that

you found and then manipulate this right here to make it look like this

and then you’re done think of it like a game

something is given to you like a card or a pair of die and and you want to start

with this right here and you can because you found the delta

in step one so you can start with this and you want to do some kind of

manipulation to make it look like that and then you’re done so you want to use

this relationship that you found so actually start part one isn’t just

necessarily finding the delta but when you’re working out to find the delta

you will usually have the steps that you need

for the relationship right there so it’s not as hard as this may seem

00:53

this seems very daunting when we write it out as a general

procedure like that so what i’m going to do is i’m going to work out an example

for you and we’ll come back to these steps and you can see the steps

for a particular example so let’s look at our first example right here

there’s our two-step process here’s our first example right here

okay so steps for proving our c is one our f of x is three x plus two in our l

is five and here’s our steps for proving this limit is five

now i need to find a delta or some epsilon given right and so i

need to start off with this right here and you know find something that looks

00:54

like this and so let’s let’s do that so nice clean board here

and we’re gonna um you know start off with f of x so here’s my f of x

so let’s start off right here f of x minus l

is less than epsilon and so i’m going to start off here with

f of x 3x plus 2 minus 5 is less than epsilon so so this is step one

let’s write that down step one is find delta

find delta so i’m going to start off with this right here

and i’m going to work to find the delta now right so to find the delta

remember to find the delta we have to um use this right here that’s why i’m

starting off with this right here and make it look like this and then

whatever that shows up over there will be our delta

so let’s start with this and try to make it look like that

00:55

x minus a c and a c has to be in our case

the c is a one so i need to make it look like

the x minus c is less than delta i need to make it look like that

but we have this so how can i go from this to that

it’s the first question find the delta and i start with this

now how can i get an x i have a 3 in front of it i need a 1 in front of my x

right so first off i’m going to just simplify this this is 3x minus 3.

and then i’m going to factor a 3 out and now i’m going to

divide both sides by three [Music] so we started with this right here and

we made it look like x minus c is less than some delta

00:56

and so now we know what the delta is this is our delta right here

so i found my delta and i showed this work here on how i found my delta

but remember this is all just scratch work so we started with the f of x minus

l which is what we want to prove to end with

and so i’m going to start with it and try to work backwards

to figure out what my delta is but now that i’ve found my delta my delta is

simply epsilon over 3. so now that i’ve found my delta

now step two is we can write a proof so step two will be to write a proof

write a proof [Music] so of course my proof won’t contain

the writer proof statement so box that off right

so here’s my proof now that i have my scratch work right here and my

writer proof right here we’re trying to prove that the limit as

00:57

x approaches one at three x plus two equals five

write a proof of that so here we go um and we wanna show you what we wrote

before write a proof so we’re going to start

off with epsilon grading zeros given and we want to start off by assuming

this and we want to use this relationship that we found to end with this

and here’s the relationship that we found you know by working through this

work here so here we go let epsilon great be greater than zero be given

so epsilon here is any real number greater than zero and that’s given to us

so if i’m going to say uh set delta to be epsilon over three so i’m giving away

the end of the story epsilon uh delta is epsilon over three i found my delta

00:58

and here’s what it is and now i’m going to show that that delta works

here we go if um x and c which is one is less than the delta which is epsilon

over 3 then so now i’m here and now we want to

get to the ending over here so how do we do that so if this is less than

the x minus you know one here [Music] then i can say the f of x minus

l so here’s the f of x minus the l which we know is equal to the three x

minus um minus three and that’s equal to three times

x minus one and now we know that this is less than

00:59

i mean we’re assuming that the absolute value of x minus one is less

than the epsilon over three so now i write less than and then i have a three and

then this part right here by my assumption is epsilon over three

so epsilon over three and that’s equal to epsilon so we just showed that

f of x minus l is less than and we have less than the epsilon and so

we showed some intermediate steps these are equal and these are equal but

here i’m using less than because this right here is less than right here so

here’s the proof right here let epsilon be greater than zero be given

i’m gonna set my delta to be epsilon over three

so i showed there exists a delta i’m saying what it is right here very clearly

no matter what epsilon is for all epsilon grading zero

01:00

i showed there existed delta here it is such that if this is less than epsilon

over 3 this is less than the delta then f of x minus

l is less than the epsilon so this is the proof right here

this meets the exact uh definition of the definition of a limit okay so

there’s our first example there and finding the delta is the first step

and then writing out the proof is the second step

so let’s do that process again and see if we can get it so for next example

i’m going to work with a negative c sometimes that trips students

up so here what is changing [Music] we want to find our delta so i have f of

x minus l that’s where i’m starting from is less than epsilon

01:01

and let’s see how this changes so this time our f of x is 7 x plus 12 minus

now the l is a minus 2 so i have to be careful this is gonna be minus two in

here that’s less than the epsilon okay now i know this is going to be the

ending part of my proof because that’s f of x minus l is less than the epsilon

but to make this argument work i have to find the delta

so what will the delta be in this case so here we go let’s say here this is 7x

and this is 12 plus 2 so that’s a 14 so that’s plus 14.

and we can factor out a 7 here so we can say or we can just divide both

sides by 7 so we’re going to get 7 and then absolute value of x

plus 2 is less than epsilon and then now keep in mind though that

01:02

what we really want is x minus the c so i could write it with a plus 2

but here’s what i’m thinking in my head it is it’s x minus a negative two

so that’s what i’m thinking even though i wrote a plus two so you could write

plus two here or you could write it like this but it needs to be x minus the c

so now i’m going to divide by seven and we have our delta here this is our delta

whatever we get right here we have delta as long as you have x absolute value

x minus c and then that can be your delta here now remember your delta can do

depend upon the epsilon but it cannot depend upon x so this is our delta here so

in this case it’s going to be epsilon over seven so i’m going to go and write

a proof now let epsilon be given no matter how how small epsilon is

01:03

let f be given i’m going to set the delta to be

epsilon over seven so no matter what epsilon you give me

i’m coming up with a delta so delta exists because you gave me epsilon

so if and now here we have x minus c in our in our case the c is minus 2

so that says x minus minus 2 and then absolute value

let me write a little bit better x minus -2

is less than the delta which is epsilon over 7

then now keep in mind all these words little words right here matter

this is not a limit computation where you know you didn’t need any words this

is a proof so it definitely needs words so i’m starting off by

you know writing what epsilon is and what delta is

01:04

and i’m using an if then so i have an implication going here so if this right

here is holding then now we have to come down here and

see if we can fill in the details here we go then the distance between

f of x which is the 7x plus plus 12 or right here so 7x

plus 12 minus the l which is the minus two there um yeah minus two

so that is equal to seven x plus fourteen which is equal to seven

times x plus two which is equal to seven times the absolute value of

i’ll put it over here seven times the absolute value of x minus a minus two

and that is less than this right here is less than

01:05

seven times remember we have a bound for this

right here so this absolute value of x minus the the the distance between

x and minus 2 is less than so this will be epsilon over 7

and this is just epsilon so there we go no matter what epsilon is given

we found a delta such that if the distance between x and c is less than delta

then the distance between f of x and l is less than the epsilon so

we proved this limit is in fact minus two using our epsilon delta r

uh proof right here so here’s the proof right here [Music]

01:06

okay so let’s see if we can do one more [Music] so in the last two examples we

let’s take a look at the last two examples real quick in this example

we let our delta to be epsilon over three and then this um and then in the next

example we let epsilon be uh we let delta be

epsilon over seven so what do you think the delta will be in this problem

i’m guessing the delta will be epsilon over five so in other words the point is

is that after you do enough practice it takes you less and less time to to go

through the process of writing you don’t need as many drafts

in fact maybe you don’t need any scratch work or brainstorming at all

so this is what it would look like if you’re just going to write a proof

and you already know how to write the proof so use an epsilon delta argument

01:07

to show that so i can write a proof proof let epsilon be great epsilon greater

than zero be given set delta to be the epsilon over and i’m gonna go with the

five here as my as my value for epsilon but that’s my value for delta

so let epsilon be gradient zero be given set delta to be epsilon over five

okay so if so if um 0 is less than the the distance between

the x minus the c in this case is four is less than the delta epsilon over five

then now here we go we’ve got some work to do here

01:08

the f of x which is 5x minus 2 minus so that’s f of x minus the l 18 is

equal to what is it equal to 5x minus 20 which is equal to 5 times the absolute

value of x minus 4 which is less than the epsilon over 5.

so 5 is the same and then this part right here the absolute value of x minus 4

is less than epsilon over 5. and so that’s epsilon over five

and so this is just epsilon so if this is less

if this right here is is true then all of this is true

01:09

and so that’s really all you need there for your proof hence

the limit as we approach four the five x minus two is eighteen [Music]

so let epsilon be given there exists a delta such that if the distance between

x and c is less than delta and and and x and z are not the same so if that

happens then the distance between f of x and l is less than the epsilon

and we showed that the distance between f x and l is less than epsilon

by doing a little bit of by doing a little bit of work we just simply said

that’s minus 20 we factored out a five and here we used our hypothesis

that the distance between x and four is less than the delta which is epsilon

over five so we showed for all epsilon given there exists a delta

01:10

such that this implication holds here and that’s all we need to

prove that the limit holds here okay so now what happens though

if we have a quadratic so now we’re going to need to do some more draft

scratch work whatever you want to call it so um we’re going to start off with

[Music] f of x minus l needs to be less than epsilon so this is

step one is to find the delta so i’m going to start off with where i

need to end and see if i can build the connection what is the delta

to make the argument work so in this case we have f of x

and then minus the 4 the ls4 is less than delta is less than epsilon now

01:11

we can write this as x minus 2 and then x plus two

now you know x is approaching two so i need to start with this and i need

to make it look something like this so this is where we’re starting this is

where we’re ending when we write the proof we’ll start here

in our hypothesis and we’ll get here if we can if we can find the delta

so let’s see now what’s happening well we have x minus two here and that’s good

but we we don’t have you know what we cannot let me show you

what we cannot do sometimes that’s informative

so we can say here oh let’s just divide both sides by

the the absolute value here now i found my delta

right so now i have looks like this doesn’t it

and this right here is my delta so that’s what we did when we had a

linear function up here um it was able we’re able to make the connection pretty

01:12

easily but now this is involving x and so this you know

this is not going to work so we need a a delta here

that can only have an epsilon in it it doesn’t it can be a number

delta but in fact delta can have at most an epsilon in it delta

cannot have an x in it so i cannot divide both sides by that’s the value

and and get a delta that way but there is another way we can get a delta [Music]

so you know when we’re looking at approaching 2 here for this limit right here

means we’re pretty close to two so you know if we’re if if x is in this interval

right here if x is in one to three what’s happening so if x is between

one and three what what is this right here doing

01:13

so then the absolute value of x plus two is less than five in that case right

because you know you got the three and you got the one so it’s less than five so

what i’m going to do is say let delta to be equal to the minimum of the

epsilon over five which is what what what we have here and one and so [Music]

if i choose this is my delta right here then we will be able to

write our proof and so i’m going to choose the minimum because

whenever this is holding right here what we have for

this right here so this will be x minus two um x plus two and

we can say that this is less than and then this right here will be

01:14

epsilon over five and then we’ll have a five

because this is true right here and then we’ll get an epsilon right here

so this will be uh epsilon over five we’ll use a five here and then we’ll be

able to get the the epsilon right there and so now you know we can um

write out our proof here so let’s see let’s see

let’s write out right out of proof now that we know what the

delta will be okay so here we go so proof let epsilon greater than zero be given

and i usually like to say right up front what my delta is

so set delta in other words let delta be the minimum of one and

01:15

epsilon over five and that would be my so whatever epsilon

they give me if they give me point zero zero zero one and i divide by

5 that would be the minimum in that case if they give me epsilon is say 500

then 500 over 5 then 1 will be the minimum there [Music] okay so if

um x and c are not the same so greater than zero

so we have x minus and what’s our c two so if that is less than the

delta right here because it’s a minimum i’ll just say delta here

then and now i’m going to start with f of x minus l and i need to say

blah blah blah less than blah blah blah less than epsilon

you know i just need to have some work in here and get less than epsilon

01:16

and then i know i’m done so this is just following the structure of the

definition for all epsilon grading zero let’s get

for all epsilon greater than zero there exists a delta such that if this holds

then this holds so using this delta right here and this

hypothesis right here using those two things and that epsilon is given so

using those three things prove that the distance between f of x and

l is less than epsilon so let’s see how to do that so in this case our f of x is

uh x squared and then minus the l and then now remember our scratch work

this is x minus two x plus two and so then we can write this as

less than and then the epsilon over five and then times the five and so we’re

01:17

going to get five here so it’s really making use of the less

than here and the way that this is set up as the minimum right here

so we proved that f of x minus l is less than

the epsilon there’s the less than right there [Music] so

that that is the proof there and let’s look at one more example here

so let’s look at our scratch work first here and so this time let’s write it out

differently let’s write it out as not necessarily just such a short proof

but let’s write out our whole thinking process this time

so i’m not going to just write out proof this time

i’m going to also write out our thinking process and

so you can see both types so here i’m going to say

01:18

given epsilon greater than zero we need delta greater than zero such that if

the distance between x and c is less than delta then the absolute value of

f of x and l is less than epsilon less than the epsilon

so i’m writing out what we need so we need to show there existed delta

such that if this holds then that holds now you know simplifying

01:19

so x squared minus 4 here is less than epsilon whenever

this is holding x plus 2 is less than delta

so just simplifying this right here so we need to have this hold whenever

whenever we’re assuming this right here is holding so notice notice that

if the x plus two is less than the one then you know what does that hold here so

then we have minus one x plus two less than one here

taking the absolute value off so you know we have here

minus five less than minus two you know to get to a minus two here

01:20

we’re just gonna you know subtract off of four here we get minus three so that

the distance here is less than five so we need this distance right here

um because we’re going to have x squared minus four so this is

5 here and so in this one right here take delta to be the minimum of

1 and the epsilon over 5. and so then we can run through our

argument here so take delta to delta to be the minimum of one or

epsilon over five so now we can run through our argument here [Music]

then zero is less than x plus two less than delta

01:21

implies that x minus two is less than 5 and x plus 2 is less than epsilon over 5

so we have both of those if this holds then we have this one’s less than 5

and this one’s less than epsilon over five so for the big finale

f of x x squared minus one minus the l which is three

is equal to as we said above this is the x squared minus 4 which we

can factor as x plus 2 times x minus 2 [Music]

and that is just simply x plus two x minus two and now using our

inequalities here this one’s less than five this one’s

less than epsilon over five so this will be you know

01:22

epsilon over five right here and then this will be five x minus two

is less than five and so this is epsilon there

and so this is showing some of the insight in terms of

in terms of the proof so this is a this is a proof also

it’s just a lot more wordy and so you know let epsilon be given

we need to find a delta and so we’re showing our thinking

we’re not just saying let epsilon be given here’s what delta

is and now let me prove that and now let me just show you that

that that delta works so we’re showing some of our thinking here

let epsilon be given we need to find a delta such that

if then and so now we’re looking at the f of x minus l we’re simplifying that

and we’re coming up with some some uh because we have a factor it’s a

quadratic so in this case we can factor it pretty easily so

01:23

we have x minus x plus two that we’re worried about

um and and so we have x plus two but then we get it to the x minus 2

because we also have that factor so no matter which factor we have here

we have a bound for it and we can come up with f of x minus l that distance

is less than the epsilon right there so that would be

a perfectly perfectly well formed proof even though it’s a little bit wordier so

some people like the wordier version some people like the

strict ears let me not show you any of my

scratch work let’s just do the proof so we did that last example

and so here’s one where you would see more of the thinking involved

okay so um next topic [Music] okay so for our next topic we’re going

01:24

to work out some limit theorems and now for these limit theorems here

um i’m not going to just go straight for the theorems here i’m going to

there’s going to be some crucial steps in these theorems

that are going to need some kind of help so whenever you want to provide some

help for some big theorems that you’re about to talk about

you often find limits first so this is called the triangle inequality

now in some classes you might call this a theorem like in say for example

precalculus but here it’s going to be something to help us

so in other words i don’t want you to think that this right here is a limit

theorem this is not a limit theorem this is something to help us

prove that a limit theorem is true it’s called the triangular inequality

so i just want to don’t don’t lose anybody at the word limit limit just means

01:25

this is something we’re going to prove is true and it’s going to help us

prove another theorem so it’s a helper all right anyways

when i’m looking at this proof right here i’m going to

work with the squares of this right here so let’s see how i do this here

so i’m going to look at the square of the absolute value and the reason why

is because this is a lot easier to work with

in fact i can remove the absolute value now if you put square on it

so if you take the absolute value and then square it

well it’s going to be the same you square it

all right and now i’m going to multiply this out that’s x squared plus 2x

plus y squared now oops got my y now it may happen that x and y are both

01:26

negative in which case this will be positive it

may happen that one is negative one is positive

um so if i were to go and put absolute value on here

i could not say equals we cannot put equals here oops x times y

so we cannot put equals here if i want to just put absolute value around the x

and the y and i’ll show you why i want to put

absolute value around the x plus y’s in fact i want to put absolute values

around here also now here it’s not a problem

because they’re already squared here they’re squares they’re squares here so

x squared is the same thing as the absolute value of x squared

for example if i have 3 squared it’s 9 and if i have the absolute value of 3

squared is still nine and if i have negative three squared is nine

01:27

and if i have the absolute value of negative three squared

it’s still nine so it doesn’t matter if you have absolute value

sorry i was doing that off screen there um

if you have three squared is nine and if you have absolute value

so i’m trying to explain are you okay here you should be comfortable here

because if you take the absolute value of something but you still square it

these are going to be equal to each other so if i have three’s positive right

or if i have if i minus three here it doesn’t matter if you have absolute

value around the minus three or not because you’re going to square it in the

end so these are the same here and these are the same here but this is

not the same here these are not the same because x could be negative

and y could be positive and so but this is always positive here so

01:28

for example for example if x is say a minus three and a y is four

that’s not the same thing as the absolute value of minus 3 [Music]

and the absolute value of 4. if i put the absolute values on each one

put the absolute value on the minus 3 and put the absolute value on the four

this right here is right uh you know minus this minus 24 and this right here

is positive 24. because this is just two and then the absolute value

and then four this is 24. so you can see they’re not the same

so if i want to put absolute value on these x y here i have to say less than or

equal to you know this one with the absolute

values here might be greater than this one

all right so long story short i’m going to put

absolute value here on this one and this one

and that’s fine but i’m going to put absolute value on this one right here

01:29

x times y but it may get larger so now i have each one here

so i put the absolute value on the product or if i put the absolute value

on each one it’s still it’s still going to be all positive right in here so

it doesn’t matter if you multiply first and then take the absolute value

or if you take the absolute values and then you multiply in either case those

are the same now this is the absolute value of x plus the absolute value of y

all with the squared on it because what does the squared here mean right so

it’s absolute value of x times the absolute value of x here and

it’s absolute value of y times the absolute value of y here and

then it’s the mixed terms absolute value of x times the absolute value y

times two so but we have an inequality here so we have to be careful here

so you know we have this nice computation right here

01:30

but we have less than or equal to here so now we can say

um the absolute value of x plus y is less than or equal to because that’s

that’s the most we can say is less than or equal to this one right

here absolute value of x plus the absolute value of y here

so this is the triangular inequality now in terms of a triangle uh right so

you may say something like this is the leg and this is the leg and

you know you got this length plus that length

and then you know the shortest distance between

two points here you know this is going to be the longer route

around the triangle perhaps absolute value of x plus the absolute value y

so you know think of it in terms of a triangle then you you can see

01:31

that these are the sides and then this is the larger

but if you do both of these you’re going to take a longer route than

you know going between straight going between two in any case that is the

triangular inequality and that’s how you would write a nice proof of it out

is by looking at the squares first and then realizing

that you can have the absolute value i mean uh

square root of both sides there it still preserves the inequality there

all right so this will be an important uh inequality to use

in our limit theorem which we’re about to see but first we need one more lemma

and this lemma is kind of obvious but it’s important that we uh talk it out and

i don’t want to surprise you when you get to a theorem

and so i want to flush this out first so we’re going to say r is a real number

01:32

any real number but it’s greater than or equal to zero okay

so that’s what r is so r could be 10 could be zero could be 100 could be a

million now here’s an if then if r is less than epsilon

and epsilon could be any real number so if r is less than any other real number

um that’s positive epsilon is positive then our in fact has to be zero now this

actually kinds kind of sounds obvious but just to make sure

that we thought about it before we’re going to use it so here’s how a proof

would look so suppose that r is greater than zero suppose r is greater than zero

strictly greater than zero so this right here says r is greater than or equal to

01:33

zero but what what happens what’s so wrong with r being greater than zero

okay so then by hypothesis so r is less than any epsilon greater than zero r is

greater than zero so r has to be less than r

right but that clearly cannot happen r cannot strictly be less than r

so therefore we have r is less than or r is greater than or equal to zero

but then is less than or equal to zero and so r has to be equal to zero

so r cannot be greater than r so you know this is not something that can

happen here so we actually have the hypothesis here

that r is greater than or equal to zero right here

01:34

we just suppose that r is great strictly greater than zero

and saw that that cannot happen so r must be

what’s the opposite of that r must be less than or equal to zero

right that cannot happen we get a contradiction since that cannot happen

r must be less than or equal to zero and that’s just uh our given hypothesis up

there so both of these must hold this one holds from this

previous sentence this one holds by given up here

so if this if these all hold then r has to be zero so um that’s nothing too

insightful all by itself um but it’s very useful thing to prove that if you

know something is greater than or equal to zero

but it’s less than every other positive real number

then in fact it has to be zero so we’re going to use that right now

here we go so here’s our first limit theorem now that we’re

01:35

much more familiar with the limit definition by working out those examples

now if you’re still not familiar with the limit definitions

you may want to practice some examples on your own

i have some exercises at the end of this video that you can practice

but definitely want to practice you know here’s one here

x is approaching one and you got four x plus three

so by the first episode this video you know that you can easily find this limit

it’s just seven so practice writing an epsilon delta

argument you can easily just make up an easy limit that’s easy to find but you

want to prove using epsilon delta argument

and once you convince yourself that you understand what that’s

how that’s working then we can take the next level

and actually try to prove a theorem so on this theorem here we’re given two

limits here this one and this one and we’re asked to prove l equals m

01:36

now what this is saying is that limits are unique in other words if

if this limit is l this limit is m then l and m

are the same in other words there can only be one value for the limit

if you think there’s two values for the limit well you’re wrong

they’re equal so let’s go see why this is true

and the we’re going to use our previous lemma two limits to do this

so i’m going to start off by saying let epsilon be greater than zero be given

and now since the limit as we approach a of f of x is equal to l so what that’s

saying is by the limit definition for this one so there exists a delta

right epsilon’s already given so there exists a delta

01:37

now because i have two limits in this problem the limit for f of x and the

um is l and this one right here for m i’m going to be applying the limit

definition twice so i i’m going to i’m going to uh subscript my deltas

so by applying the limit definition here with the l

so there exists a a delta 1 greater than 0 such that [Music]

if this holds if this is less than delta 1

then that then it must follow that f x less than l is less than epsilon over two

now you might say where did i get uh epsilon over two

so here’s what i really did now epsilon is given

so that means i know that what epsilon over 2 is

right so think of think of epsilon as like 0.1 or 0.5

we don’t know what it is but it’s been given to us

01:38

so since epsilon has been given to us i can chop it in half

and now i know this number so i’m actually applying this limit definition

right here with the epsilon over two epsilon epsilon over two is given to us

so there has to exist a delta such that this implication here is holding

now similarly we can write the same exact sentence down again

but this time we’ll use m so since this limit right here is equal to m also

so there exists so we have the same exact argument

epsilon’s given so that means we know what epsilon over 2 is

so using this definition right here with epsilon over 2

epsilon over 2 is given so there exists a delta

this time i’ll say subscript 2 greater than 0

such that the distance between x and a less than delta 2 implies that f of x

01:39

and this is applying it to m here is less than epsilon over two

okay so since epsilon is given we know epsilon over two

we can apply the definition over here with epsilon over two

we have to get a delta such that this implication is holding

we have to get another delta such that this implication is holding

now to finish the proof we’re going to combine them together

so let delta be the minimum of these two deltas over here

so minimum of delta one and delta two and if we start off with uh x minus c

is less than delta here or not c sorry a

then and here’s where we’re going to use the triangular inequality

01:40

we’re going to start off by saying 0 is less than the absolute value of l m

now l and m for all we know might be the same

might not be the same we’re trying to prove they are the same

so to prove they are the same i’m going to look at the distance between them

and i know that the distance right absolute value so it’s greater than or

equal to 0 here now here’s a very nice trick here

so our technique so this will be l and then minus

f of x see because we have l and f of x is up here and m and f of x is up here

so we have this and this so i want to try to put some

f of x’s into my l’s and m’s so i’m going to say l minus f of x and

then i’m going to say plus f of x minus m so

all we did was we added and subtracted an f of x here

so that i can put some f of x’s in here because we’re trying to use this up here

01:41

that we know is true but we don’t have any f of x’s in here so i put them in

now remember we have the triangular inequality it says that if you have two

is less than or equal to the sum of the individuals here

so i’m going to think about this right here and this right here

and i have an addition between them so this will be

less than or equal to the first one l minus f of x plus

and then the second one f of x minus l so this is using the

triangular inequality to go from all of this here with absolute value and

i’m thinking about this right here is an x and then a plus

and then i’m thinking about this as a y so this is less than or equal to

the absolute value of this right here the x plus

01:42

plus plus the absolute value of the y right here okay so now the point is

is that we’re trying to use these statements up here that we know are correct

so this is f of x minus l right here now if they don’t match exactly that’s no

problem this one matches exactly right f of x minus m

this one’s a little bit backwards but you know keep in mind that we can

factor out a minus 1 and say f of x minus l now and then plus this one

and then take the absolute value of the minus 1 and then the absolute value of

this one so this would just be absolute value of f of x minus l

so the minus sign does not really pose a problem for us

but now we know this one is less than epsilon over two

01:43

and this one is less than the epsilon over two

we add them up we get less than the epsilon

now what we’ve shown is by limit number one that therefore l equals

to m because this right here now has to be zero so what we showed is that

the distance between l and m is less than epsilon for which epsilon or

every epsilon every epsilon um the distance between l and m is less than it

and that’s greater than or equal to zero so now this is lemma one

saying actually what we showed is that the distance between l and m

is zero that this right here is less than an arbitrary epsilon

so but if the distance between them is zero right so in other words so that is

01:44

l equals m as needed okay so there is the existence uh sorry not the existence

there is the uniqueness of limits the limit of a function

must be unique if you come up with two different numbers that you think are

different well you can run through this argument

right here apply the definition for that l apply the definition for that m you

get these deltas that have to exist you choose the minimum delta here and

then you can bound the l and m or an arbitrary epsilon so they have to

be in fact equal okay so let’s work on one more limit theorem now the first

limit theorem that we just showed this one right here

this one is actually often taken for granted among calculus students

you tend to just say oh something’s defined it must be unique

01:45

that’s certainly not true that’s something that you have to prove in mathematics

if you want to be very very rigorous about your work

to not just assume something is true that just because you have the limit is

equal to one l that all limits have to be equal to l in fact this shows

this theorem shows that all of them do but

this limit theorem here is different in the sense that

we’re going to use it all the time i mean you use the first limit all the

time too it’s just that you may not have realized it now this example here

we showed in a previous example that let’s see here what did we show

we showed this limit right here minus two x squared minus one is three

we showed that in a previous example we also showed that the limit as we

01:46

approach two of x squared is four and we also showed that

the limit as we approach minus two of seven x plus twelve we used a

epsilon delta argument to prove that this was -2

so here’s three limits that we proved is or is true using an epsilon delta

definition now what is what does this theorem say so this theorem says that

if you’re approaching c and this one’s approaching c x is approaching c and

x is approaching c then if you add up the functions and get that new function

then if you approach c well you can just add the l and the m together

so here’s an example what is the limit as we approach

minus two of what happens if we add these up

01:47

we’re gonna get x squared plus seven x and then plus eleven

right so let’s just write that out so the limit

of the f of x right what is the f of x let’s say x squared minus one so f of x

plus and then what’s the g seven x plus twelve

so this is the limit of f of x plus g of x so on one hand the theorem says

that this is the limit as we approach minus two of f of x

plus the limit as we approach minus two of g of x we found these two limits here

already this limit was a three and this limit right here we found as a minus two

so this is just a 1. so we found this one right here to be

01:48

a 3 and this one to be a minus 2 so this limit of all this is just a 1.

so this limit right here if you know if i just add them up it’s x squared

plus 7x plus 11 right this limit is 1. now how do you know this limit is one

right here well we proved this limit is three and this limit is minus two

and we have this theorem right here that says to find the limit of f

plus g which this is what is what is what this

is so i can just take the limit of the first function

and plus the limit of the second and since i already found those

so now we know this is one here and so you know we prove these limits

here using the limit definition however you want to find those limits i

can now find the limit of this one right here much much faster

because i just add the limit values together i get one it’s

01:49

it’s so using limit theorems is much faster than going to prove this limit right

here is true using a limit definition i don’t need to go prove this limit

is true using a limit definition i can do it i could say for all epsilon there

exists a delta i could prove what that delta is

i could go work out that proof but we don’t need to

because we’re going to prove this limit theorem right now it’s true

no matter what f of x and g of x are um actually

um so let me just erase this here for a quick minute i just realized

that this should be uh g of g of x here so that should say that so let me go to

the board here now and make sure i write this down so i don’t lose anybody here

01:50

so the theorem says if the limit as we approach c of f of x is l

and the limit as we approach c of g of x this should be a g of x here

then the limit of the sum f of x plus g of x is equal to l plus m okay so

i messed that part up right there that should be f of x that should be g

of x right there in any case let’s write out a proof

so this should be g of x right here as x approaches c of some function

g is m so if i want to find the limit of these two

now we can use the limit definition a limit definition to work on those

but it doesn’t really matter how you found those if you want to add them up

01:51

together just add the l and them together now this is assuming first of all

that this limit exists and it’s l and this limit exists

and it’s m if either of these limits don’t exist

then you may not be able to use this well then you cannot use this theorem here

this theorem is going to assume the limits do exist

all right so let’s write a proof now so let’s get rid of this scratch work here

so here we go so proof again before we look at such a proof you may want to

stop the video and work out some more examples

and get familiar with you know because it’s hard to

get familiar with something as complex as the limit definition

just by watching someone else work out some examples so you need to

stop the video now perhaps and get more comfortable with some more examples

01:52

but when you’re comfortable enough with the epsilon delta

limit definition then you should come and look back at this theorem here

so assume epsilon is given so we’re going to start with that why

well because we’re trying to prove this limit over here on the right

is l plus m we’re trying to prove that so

in order to prove a limit is equal to a value

using the limit definition we start off with an epsilon given now so

since the limit as we approach c of f of x is equal to l there exists a delta

01:53

let’s say delta one grade zero such that such that the diff the distance between

f of x and l is less than epsilon over two whenever the

x and c here have are not the same and less than the delta one

right so i’m applying the limit definition right here to this

l right here for f of x and l so since f epsilon is given that means

we know what epsilon over 2 is and then i can apply the definition there existed

delta 1 to get the existence of delta 1 apply the limit definition

such that if this holds then that holds now sometimes you write if then with the

whenever right so you could say something like if p then q or p

implies a q or you could say q whenever p so there’s a lot of different ways

01:54

of saying this p implies a q is if p then q or you could say q whenever p

so i chose to write it like that this time okay so now we can write the second

statement out for the g right because we’re given both of these

limits here so i’m going to say since the limit as x approaches c of g of x here

remember that should be a g of x up there in that statement

so g of x equals m so there exists a delta two now so i’m using delta one

when i apply the definition for here and i’m using a delta 2 when i apply the

definition here such that this right here holds g of x minus the

m so this is an m and this one is less than epsilon over two whenever

01:55

now you might say why do i keep choosing epsilon over two

and the answer is because i know that if i use an epsilon over 2

then in the end i’m going to add up my epsilons and get out an epsilon

as i need to so here we go i’m going to let delta be the minimum again

so delta will be the minima of these two deltas over here so delta one delta two

and assume that x and c are not the same and is less than this one

so you know if x y c is less than delta the x minus c

is certainly less than the delta one and it’s certainly less than the delta two

01:56

or equal to it but so that’s why we choose the minimum now

now what we can say is because the delta is less than or equal to the delta one

so you know we can finish the deal now so assume then

so we have the f of x plus g of x so when i look at this limit up here the

limit as x approaches c of f of x plus g of x right

so now my function is f of x plus g of x so i have f of x plus g of x and then

minus and now what’s the value what’s the limit over here it’s not just an

l or an m now it’s l plus m so this is going to be a minus an l plus an m

so this is my function for the limit i’m trying to prove over here on the right

so think of this as like the function minus

the the limit that we’re trying to find here l plus m

01:57

now this is going to be equal to we’re going to rearrange the terms here

so we’re going to say f of x minus the l and then plus

the g of x and then minus the m right so just rearranging terms and

putting them together because as you know we’re going to come up here

and and use these two over here so we need f of x minus l and we need g of x

minus m okay now here’s where we’re gonna use the triangular inequality again

we have this broken up by plus so i’m thinking about that as an

x and that is a y and then just to refresh your memory one more time

if you have an x plus a y so this is all my x here

plus and all my y this is less than or equal to

if i take the absolute value separately so this is going to be greater than if

01:58

you add them first and then take the absolute value so here we go

that’s the value of the first one f of x minus l

plus the absolute value of the g of x minus m here and now we can use

now that we got the distance between f x and l

and the distance between g of x and m we can come and use our

inequalities up here that we know must be here

because those limits are by assumption existing

and equal to the l’s and m’s so here we’re going to have now

less than for this one right here we have epsilon over 2

and this one right here we have an epsilon over 2 in so this is in fact

epsilon so we showed that the distance between this function right here the sum

function minus the sum of the limits is less than or equal to or is less than

01:59

arbitrary epsilon so that does it that does our proof there

for the sum of the limits there’s our proof there

always starts off with epsilon is arbitrary

and then we have the distance between the function and the va

limit value is less than the epsilon and we got there by applying the

definition to each one okay so we showed that for all epsilon

grade zero is given there existed delta if we assume this is true then we know

that this is true and since we can since we know this is true we can use this

and then or um it’s equal to the minimum so it’s delta is equal to

less than both of them so we know this one here is also true

02:00

so we can use this one right here also so if this is true then this is true

that’s why i’m allowed to use these down here because for this

delta it’s less than or equal to both of them

so i can use both of these over here i know both of these are true

and this says the limit definition says if these are true

then these are true and so now i can use these down here

to finish off my argument there so that’s the

broad strategy as well as the individual steps okay so there’s the limit law

our first limit law there our first limit theorem

and using the limit definition to prove it

now there are a lot more limit theorems in fact in our first video

i showed you what a lot of the limit theorems were

without proof and we just practiced finding limits

using those limit theorems and so there’s an example of how to prove

02:01

one of the limit theorems okay so next part okay so now let’s talk about some

exercises so perhaps at some point in the video you stopped and

decided to go to the end and look at some some theorems here or some exercises

that’s always a great idea um so maybe you’ve tried some of these

and gotten much more comfortable with the precise definition of a limit

and now the exercises two three and four are are limit theorems um you know you

got to start your limit theorems off in small little chunks and start trying to

build them up and you get more comfortable and more skilled

at using the definition of a limit and then we have some more so here’s

02:02

some quadratic functions here so number one you your hint would

probably be to factor and you know use the limit definition

um factor and cancel you know that sort of thing

and then exercises six and seven well six is

very similar to the one we did before we did the sum

this one is the um g of x here now i notice here that i also have a

typo on number six here so let me write out number six here it

has the same typo that we had before so you know this

exercise six right here should stay the limit as we approach c of f of x is

equal to l and the limit as we approach g of x so this says an

f of x above it should be a g and so we have this

difference of two functions here is the difference of the limits over here

02:03

so try to work out um five and then get more comfortable with epsilon deltas

arguments and then try your hand at six and seven

now if you get stuck on any of these or if you have any comments

um you know use the comment section below to

let me know about which ones you want me to get back to you on

and i can try to help you and so let’s go on now to

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02:04

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