# Using the Limit Definition (Precise Definition of Limit)

Video Series: Calculus 1 (Explore, Discover, Learn) Series

(D4M) — Here is the video transcript for this video.

00:00
[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

00:01
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
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

00:02
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

00:03
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
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

00:04
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

00:05
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

00:06
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

00:07
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

00:08
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

00:09
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

00:10
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

00:11
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

00:12
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

00:13
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

00:14
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

00:15
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

00:16
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

00:17
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
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

00:18
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

00:19
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

00:20
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

00:21
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

00:22
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

00:23
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

00:24
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

00:25
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

00:26
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

00:27
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

00:28
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

00:29
is going to represent the distance between the distance
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

00:30
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

00:31
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

00:32
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

00:33
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

00:34
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

00:35
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
and what the words mean but another point is
is that this is very useful for making computations right deriving values of

00:36
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

00:41
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

00:42
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
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
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
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 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
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
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
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
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
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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