00:00

[Music] you’ve probably heard of them but what exactly are isotopes isotopes

are mentioned in precalculus but without a rigorous definition you

may have been left wondering in this video i go through

precisely what horizontal isotopes and vertical isotopes are so in this video

we’re studying limits involving infinity [Music]

hi everyone welcome back to the calculus one explore discover learn series this

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episode is uh isotopes uh horizontal and vertical limits involving infinity

and so i’m going to start off by talking about limits to infinity and

we’ll be talking about vertical isotopes and i’m going to call this uh

short i’m going to explain how this is short-term behavior

and then we’re going to talk about limits to infinity

and we’re going to just you know talk about horizontal isotopes

and this will be long-term behavior and then stick around to the end where

we’ll go over some exercises let’s get started

all right so up first r uh limits to infinity now before we start with limits to

infinity i wanted to briefly talk about what limits are

um you know as we’ve seen them at the beginning remember at the beginning of

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calculus one uh we studied limits in the first episode

and so you know we have the limit as x approaches c

of f of x equals l and i wanted to go through briefly

what that meant or what that means uh before we introduce uh a new type of

limit here and so yeah i just wanna um go through this here and just kind of

briefly review this for a moment so let’s see if we can do that

so what does this right here mean the limit as x approaches c of f of x

right limit so if we’re going to say this limit is

equal to l what does that mean right so let’s say we have a sketch

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we have a function coming through here and let’s say right here we’re at c

and you know we don’t have to have a point defined at c

because remember limit is describing behavior

it’s not at c so it’s the behavior as we’re approaching c

so what’s the behavior of the function as we’re approaching c

now remember we can approach from two sides we can approach from the left or

we can approach from the right so let’s say i’m approaching from the

right first so i’m choosing values there’s infinitely many values

close to c and if i pick one just pick pick one then i can go up here

on the graph and compute a height and then i can choose another one

another x and go and compute a height and then i can choose another one

and go and compute a height and we keep choosing these values for c

that are closer and closer we keep choosing these values

x that are closer and closer to c but we’re choosing them from the right

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and so we ask the question what are the heights approaching

is the functional values the outputs are they approaching a value

we can see here yeah they’re approaching an l

so if we can do the same thing from the left

so here’s a number an x value on the left side of c

we go up and compute a height and then we choose another one we can go up

compute a height and we keep choosing values that are

closer and closer to c really close and as we can see here the height is

approaching some value let’s call it l now if the graph it looks something like

this where this height here was m then the limit would not be equal to l the

limit would not exist because from the right side the heights

are approaching l but from the left side the heights are approaching

another value another different height m so in that case

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the limit from the left would be m and the two-sided limit would not exist now

let’s look at the right side a little bit more detail let’s just focus on the

right side here [Music] because what if someone wanted to come

up to you and make an argument they were arguing with you

so here’s the function in here coming through here there’s the function and

there’s the height l and we’re saying the height is

l and someone else is coming along and saying no the height is not l the height

is l1 so l1 is right there and i say the height is

l and they say the the limit is l1 different from l and we have an argument

so how can we settle the argument so for example i start choosing some x values

and i’m getting closer and closer to my l now i choose something like zero zero

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one and zero zero zero zero one and these are getting closer and closer to

my my uh c here all right so let’s put c at zero now

so that we can be approaching zero so there’s my height right there and this is

zero right there so there’s my y axis so there’s a hole right there but we’re

asking the question as we’re approaching zero from the rise

so this is so i chose my c to be zero just to make the example easier

so i keep choosing values from the right of zero like

zero point zero zero one which is pretty close to zero

but zero zero zero one point zero zero one is closer

and these are closer and so these are the values i keep choosing for x

and i keep getting outputs here i get an output i get an output

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i get an output and my outputs look like they’re getting closer and closer to

l so question is who’s going to be right is it l1 or l so i say the height here

is at l and they come along and say though the height is l1 and it’s a

little bit of a different value than my l and i want to

argue to them no the limit is l right so at this point we still don’t

have a precise definition of what a limit is so how could we argue that our l is

more accurate or right in their l one is wrong well we can keep going on the x’s

they went all the way to 20 places they went all the way with twenty zeroes

let’s say they went all the way to twenty zeroes

there’s twenty zeros in here and they got a number

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really close to l1 they think their limit is l1 in fact i’m gonna go to

thirty decimal places i’m gonna go i’m gonna use thirty zeros

so i have thirty zeros in here and minus l and so we can keep choosing these

x’s closer and closer to c eventually we’re going to get so if

their l one is different than my l there’s going to be some difference here

in the height right and another if their l1 is the

same as my l then then they’re the same but if they choose something else either

above or below there’s going to be some difference in height

if you keep choosing the c closer and closer to zero

eventually you’re going to get something inside that

l1 and l2 something in that height and then in that case they’ll have to

concede yes the limit is not l1 then they may argue

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ah but the limit is l2 so then l2 might be closer to my l

but i can keep choosing x values that are even closer and closer

maybe i choose 3 000 zeros in here and then a one

so that is really close you might even say extremely close

to zero so then i would get between the l2 and l and then they will have to

concede ah it can’t be l2 maybe it’s l3 so we can keep playing this argument

over and over again eventually you’ll realize that you need

a formal definition of a limit but i think this process illustrates um

you know if you go from ten zeros in here to twenty zeros in here to thirty

zeros in here you’re gonna get a lot of you know a lot more

uh closer to the value l if the limit is in fact l okay so

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in an upcoming episode we will be giving a formal

meaning to the limit but that is a kind of intuitive idea there

so in order for a limit to exist there must be an infinite number of

real numbers in the domain of f in other words we need to be keep

we need to keep choosing being able to choose something

that is closer and closer to c so remember the limit is a process it’s a process

where we describe the behavior of the function

so it’s also important remember that we’re choosing numbers

x that are close to c but we never choose actual c

c c is not in the domain and the whole idea is that we’re not interested in

what’s happening at c but we’re interested in the behavior as we approach c

what’s the behavior of the function not what’s happening at

all right so the main idea is to study the behavior of the function

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around a point in space right so um we worked on this in a previous

exercise so i wanted to kind of talk about this right here again

this limit right here we’re approaching zero so if we look at

the function here one over x so remember what that looks like

like this right here we have a vertical isotope right here at x equals zero

we have a horizontal isotope right here and y equals zero

so the graph is some shape something like that

there’s a vertical isotope here there’s a vertical isotope at x equals zero

we have a horizontal isotope we have a horizontal isotope

so what we’re going to say here is that the limit as we approach

zero from the right as we approach zero from the right

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the function is becoming unbounded so one over x here is positive infinity

now what this means is is that as i keep choosing x closer and closer to zero

the function just keeps growing for example when i choose

x is one the height is one when i choose x is one half

what is one over one half well one over one half is two

so now the height is that two and if i choose one over three

now one over one over three now the height is three

so if i keep choosing one over ten then the height will be ten

if i choose one over a thousand then the height will be a thousand

so the closer and closer we choose values closer and closer to zero

this is growing right here and it doesn’t stop growing

now someone may come along and say yeah it stops growing

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in fact i found the last point on it it’s at [Music] it’s at ten thousand

yeah i found the i found the height it’s ten thousand for this function right

here the height is ten thousand and then you’ll look at them and say

nope i can choose one over a hundred thousand

and that’s closer that’s close to zero from the right

and the height will be even greater than

that in fact it’ll be a hundred thousand so if someone tells you the height has

stopped or there’s a bound on the height you’ll just look at them

and give them an example where it’s not true so this is

positive infinity now we can run to the same argument down here

the limit as we approach zero from the left of one over x

is minus infinity so the reason why is because now we’re choosing values to the

left of zero which means we’re choosing negative

values so for example i might be choosing negative point one

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or negative point zero zero one but the point is that these are negative numbers

down here so one over a negative will be negative but it’ll still be growing

it’ll be growing downward and so again if someone says i found

where it stops it’s minus 1 million and then we just say no i found a number

even closer to zero from the left and that surpasses it it’s way down here

way down here way way way way down here so it’s never going to stop and the way

that we symbolize that that this growth just keeps falling

is by saying minus infinity here okay so these this is what i call

short-term behavior we’re looking at short-term behavior we’re looking at the

behavior around a point zero and we can see limit from the right

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limit from the left are not the same and so we’re going to

say the two-sided limit here does not exist okay so this is all an example of

short-term behavior by looking at this function and asking

how is this function behaving around zero well to the right of zero it’s blowing

up and to the left of zero it’s just decreasing without bound

so now let’s contrast that with long-term behavior

so we’re also going to be studying long-term behavior in this video

so long-term behavior is what’s happening as x is going to infinity

so this is a different type of behavior so this is the same function that we’re

looking at but now we’re asking a different type of question

instead of looking around a number like like we looked at zero

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you could look at the one or two you can look around any any number you want

but this is asking what is the long-term behavior

in other words here’s one here’s two here’s three one two and three is really

not long-term behavior though what is long-term behavior long-term behavior is

way out here let’s say this is four let’s say this is five six seven

eight it’s not really long-term behavior either

long-term behavior is way out here all right so as soon as you start

putting tick marks down you’re gonna say that’s not

long-term behavior well long-term behavior is

really long-term behavior okay well it’s just long-term behavior

so what does that mean what is the long-term behavior of 1 over x

so x is 10 x is a hundred x is a thousand you know what’s going on here

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so if i choose for example one million what’s happening here for example what

is one over one well we get one again what is one over two we get one half

what is one over three we get a third what is one over ten

what is one over a hundred what is over one over a million [Music]

so whatever you choose for your x value you’re choosing something positive

one over something positive will still be positive

so the heights here will always stay positive

they will never cross you’ll never get one that says ah

here’s a number down here it’s at minus one

you’ll never be able to get a height of negative one for example

because that will give us an x value right here

and for that x value i’ll do one over x but it won’t be negative so there’s net

00:18

no way it will be able to cross this axis here this y-axis

y-axis uh sorry this x-axis there’s no way to

cross it but we’re getting closer and closer to zero as you keep choosing

larger values for x and you keep going out to infinity so

you’re just going to be getting numbers so close that you can’t even well

[Music] maybe someone has said i found the smallest one there

it gets and that height here is one over a thousand

and it doesn’t go beyond that it doesn’t

get closer to the x-axis than one over a thousand

and someone will say no i’ll choose a hundred thousand

and then one over a hundred thousand is even closer

and i’ll say that’s the last one and then someone will come along and say

ah one million that’s even larger remember we’re going to infinity so we

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can choose any large number that we want so one over a million

if i go out far enough and go out to one over

a hundred million that’ll be even closer so this is just getting closer and

closer and closer to the x-axis but it never crosses

so it just looks like it just getting closer and closer and closer and closer

and closer so this is an example of a horizontal isotope

and we can make the same argument for the other side

however for the other side over here now we keep choosing x’s to be

negative infinity i mean a negative values so negative 10

negative 100 negative a thousand so whenever you do

one over those numbers now you’re going to get negative outputs

so this is going to be um for example if i choose say

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negative negative one the height will be

negative one right here what if i choose a negative two

then the height will be negative one half so we keep choosing values

negative one over ten negative one over a million

so this is getting closer and closer to zero but

all the values the output stay at zero i mean stay negative so all the values

are negative but it’s getting closer and closer to

zero so this limit right here is zero and this limit right here is zero

so that limit is zero and this limit is zero and

we have y equals zero is a horizontal asymptote i’m going to abbreviate

horizontal isotope is h a and we argued a minute ago that

x equals zero is a vertical isotope so i’m going to abbreviate vertical

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isotope with va so on this simple example right here one over x

we saw an example of uh wait a second um yeah zero okay so

this is an example here this simple function right here

um we can talk about short-term behavior we can look at vertical isotopes

we can look at long-term behavior and talk about horizontal isotopes

so this function right here has a lot of interesting behavior

it’s a very simple function but it illustrates

a lot of behavior here all right let’s go on so we’re going to say f decreases

without bound as x approaches 0 from the left which we saw from the diagram

and f increases without bound as we approach zero from the right

and so the two-sided limit doesn’t exist all right let’s look at this example

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here now this one looks more complicated than one over x

so on this example here i’m looking and i’m so we’re going to be computing

the limit as we approach two now when i’m looking at two there

i’m also looking at the denominator and i’m thinking to myself

the denominator is going to be zero i probably have a

vertical isotope this is something that you might have learned in calculus one

or i’m sorry this is something you might have learned in precalculus

if the denominator is zero we’re going to have a vertical asymptote

is that true well let’s look at the function here we get a 2x to the third

plus an x squared minus 16x minus 12. so can we factor that

and the answer is yes we can factor the numerator it’s

00:23

x minus two times two x squared plus five x minus six

and we can factor the denominator it’s x minus two times x plus two

so if we factor the numerator and we factor the denominator

now we can see that we can cancel the x minus twos [Music]

and we end up with two x squared plus five x minus six over x plus two

and we can find this limit here by substituting in two

this is a rational function it’s continuous on its domain

its domain is all real numbers except minus two

so if we’re approaching positive two we can simply find this limit

by using continuity we’ll plug in two everywhere

so when we plug in two everywhere on the bottom we get four

and on the top we get what um we get um eight plus

00:24

ten minus six so we’re gonna get 12 over four we get three

so this limit is 3 actually so if we were to go sketch this function right here

what we’ll see is that there’s a hole at 2 and not an isotope of 2. so remember

just because the denominator is zero that does not mean you have a vertical

isotope you could have a hole or you could have a vertical isotope all right so

now if this function right here factored and it was x minus one

if if that was the factorization of the function and now we approach two

now we’re in a completely different case because the numerator would be non-zero

but the denominator would still be zero in this case we would not be able to

cancel the way the x minus twos and so this limit is far

00:25

different than the than the one that we’re given

so this one right here would have a vertical isotope at x

at two and we’re going to talk about these type of limits right now

so in this in this example we’re just going to simply factor the numerator

factor the denominator cancel the x minus twos and plug in the three

all right so that last example was just to help us remember that

just because the denominator is zero doesn’t mean you immediately have a

vertical isotope alright so um now let’s talk about the other case

infinite limits are used to describe unbounded behavior of a function near a

given real number which is not necessarily in the domain so

they’re useful for describing vertical isotopes

all right so here are the definitions and we’re going to say that a function

00:26

is defined on both sides of c and not necessarily at c c it may have a

hole it may have an isotope but the values of f x can be made arbitrary large

so the limits going to infinity means the values of the outputs are growing

larger and larger and larger the larger and larger and larger would

you say arbitrarily large and similarly are arbitrarily large negative

by taking x sufficiently close to c so let’s just make sure everyone

is on the same page here so if we’re looking at this limit right here

and let’s just fix one of the sides here say we’re going to positive infinity

00:27

here what is this right here mean so intuitively we’re approaching some c

so let’s say we’re right here at sea [Music]

and if this limit holds i’m going to make an isotope here

we’re right here at c and we’re going to approach positive infinity

and so when we came along and say no this is not positive infinity

this is equal to some l right and here’s the l right here there’s the l

there’s the cap on it it’s not infinity well we’re approaching from the right so

i’ll choose a value even closer to c from the right

to get above that cap that they they said it was they said it was l

i’ll choose a value even closer i mean even even greater so it’ll go even closer

and then if i claim that that’s the cap then someone else can come along

00:28

and choose a value even closer to c and get up even higher and if if no one

if no one can cap it if it just keeps growing

then the limit is positive infinity all right so that’s the intuition behind

that one there it just keeps growing if someone tries to stop it

we just choose the value even closer and it grows even bigger

all right so we can do the same thing for the minus infinity now

so if we have a sketch we have us we have a isotope right here c

the graph is falling and we have a x equals c here

and so this limit right here what does it mean we approach c from the left

00:29

right what does this mean that is a minus infinity so that means the growth is

unbounded below in other words if someone tries to bound it and they say oh

it stops right here it’s some l i found the l

that’s where it stops and then someone else can choose a value

even closer to c if they can do that and then we’ll get another l1

and then someone can come along and choose a value even closer

and there’s infinitely many c’s here to choose there’s infinitely many x’s here

to choose to the left of c there has to be infinitely many of them

to choose from so we can keep finding that there’s no bound

and so it just keeps going down it just keeps decreasing like that

so then we say if that if that all happens then we say the limit is

minus infinity there’s no bound someone tries to bound it we’ll just

00:30

choose one closer and go down further if that always happens it’s minus infinity

okay so [Music] ready for our first theorem here these

theorems will help us work out problems faster so if n is a positive integer

this limit here is positive infinity uh so we’re assuming that a is a

positive real number and that n is even so we’ll look at that here in a second

if n is um odd positive but odd if the power there in

in the denominator is a positive odd integer

then the limit might be plus infinity or it might be minus infinity

depending upon which side c is approaching for approaching from the

right it’s positive infinity for approaching from the left

it’s minus infinity um what happens if a is negative so at the

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first line i said let a be a positive integer right

so um what happens if a is negative well if a is negative then this will be

negative right here on top this is a negative number so then this

would be minus infinity and then this would be minus infinity

and then this would be positive infinity right here

so that’s what happens if a is negative what if a is zero

well if a is zero here then this this limit right here zero and

uh sorry no if a is zero um yeah then the limits raw zero okay so

let’s look at that um a little bit up close so do some examples here so

what’s an example of this right here is a positive real number and let’s

00:32

approach c let’s say c is a two and a is a positive real number

so let’s say five so what is this limit right here if i say x minus two and then

n is a positive even integer so i’ll say six power

so there’s the sixth power right there so what is this limit right here

according to this theorem here a is a positive real number in fact it’s

a real number let’s say 5.2 and then we’re approaching so we have a

x minus c and c is two so we have x minus two and then we have to sum even

power here so positive even power so six so this limit here is positive

infinity so [Music] now it may happen that you forget the

theorem maybe you don’t even believe the theorem

so how would we find this limit out without looking at the theorem

00:33

so i would look at it on both sides so i would look at it from the right and

from the left so what are the one-sided limits so first of all

when you’re looking at approaching two we have a non-zero

over a zero here we have the same case we have a non zero five point two is

non-zero over zero so whenever you have a non-zero over

zero both of these are going to be some infinities

the reason why is because the numerator is staying fixed

but the denominator is getting smaller and smaller and smaller

because we’re approaching two so two minus two that’s approaching

zero so the denominator is getting smaller and smaller and smaller but the

numerator is fixed so if that happens then what the whole

expression do is growing larger and larger and larger

00:34

and the same thing for down here just because we’re approaching 2 from the left

the denominator is still approaching zero

that’s still approaching zero down there and the top is fixed

so fixed over something that’s growing smaller and smaller and smaller

the whole thing is growing larger and larger and larger so we’re going to get

some infinities the question is is it plus infinity or minus infinity

so to determine the sign i just look at two as we’re approaching two from the

right now two from the right means we’re a little bit greater than two

not very much but we’re greater than two so what’s greater than

two minus two well that’s positive right like for example two point one two

point one minus two are two point zero zero zero one minus two

no matter what you choose down here in fact because this is an

even power right here this denominator is always positive

even if you got a negative number out right here the sixth is going to take

00:35

care of that it’s going to make it all positive so

this is positive and the same thing down here the sixth power is gonna

is gonna dominate it’s gonna make the whole bottom positive

so these are both positive infinity so that’s why that was a positive infinity

right there so we can make that whole argument

all over again with any a that we choose as long as a is positive

if a is positive these are all positive infinities here

in fact it doesn’t matter for approaching two or four approaching c

if we look at c from the right or c from the left

so we’ll make the exact same arguments that we just made a minute ago

how do we find the two-sided limit as we’re approaching c

well if we approach c from the right or we approach c from the left

00:36

when we approach c from the right here we approach c from the right the

denominator is going to zero because it’s going to be c minus c

the denominator is getting going getting smaller and smaller

but the numerator is fixed it’s a fixed positive number

so that whole expression is getting larger and larger and larger

and the six is making the whole thing positive numerous positive denominators

positive so it’s positive same thing here the a’s positive the

numerator is positive we’re going to positive infinity now it

doesn’t make a difference about the six either

as long as it’s an even power here if it’s an even power here

then the denominator would be positive this would be positive a is positive

we’ll have positive over positive both of these will be positive infinity

this will be positive infinity now if a is negative

base negative then the numerator will be negative and we’ll have a negative over

a positive in this in his odd sorry n is even [Music]

00:37

so we’ll have a negative over a positive so now

for this case now all these will be negative here okay so now for the next part

[Music] what about if it’s mixed so let’s make an argument again

let’s look at an example so we have something like

uh we’re interested in the limit as we approach

let’s say two again and this time we’re looking at let’s say 5.2 again

and we’re looking at x minus 2 and what can the power be n is a positive

odd integer so let’s say third power so how would we find this limit here

well i’m gonna look to the left and i’m gonna look to the right

00:38

so there’s the right limit and here’s the limit from the left

what are these limits so the first thing we notice here let’s look at the

one from the right first the first thing we notice is that we have a non zero

over zero so as soon as you see a non-zero over zero

the growth is going to be unbounded if you have zero over zero

that’s a completely different case but if you have a non-zero

over zero that means this limit is going to be some infinity

so this limit over here is either going to be some infinity

or it’s not going to exist remember the two-sided limit can only exist if these

agree if these are the same infinity all right let’s see here

if we approach two from the right that means we’re choosing x values

little bit greater than two for example two point one two point zero zero one

00:39

and so on but the point is is that two point one minus 2 that’ll be positive

so the these values are a little bit greater than 2

so the numerator will be positive now positive to the third power is still

positive so the whole expression will be positive

numbers so as you keep approaching two from the right you keep getting out

positive numbers the numbers keep growing larger and

larger and larger but they’re all positive numbers

so this is going to go to positive infinity

now this one is a little bit different here because we’re choosing 2 from the

left that means these numbers are a little bit smaller than 2

for example 1.9 1.9 minus 2 right that’s a negative number

and a negative to the third power is negative so we have positive over negative

so these numbers do keep growing larger and larger and larger

but they all have negative signs in front of them so you might say they grow

smaller and smaller and smaller so these do not agree so this one right

00:40

here does not exist [Music] okay and so this is the way that you can see

these two limits over here on the left one is positive infinity and one is

minus infinity and there is really nothing special about the two

what’s important is that we have a non-zero over something that’s going to zero

in that case we know we’re getting it is some infinities

and then you have to figure out which is the right one now to be honest

i don’t remember these theorem like like i don’t like have it memorized

whenever i want to compute a limit like this

i just simply see that i have a non zero over zero so i know it has to be some

infinity and then i look at what i’m approaching

and that will tell me if it’s positive or minus infinity

so that’s usually the way that i think about things

00:41

let’s go on to the next uh thing here let’s look at this um limit here

so here we go we have the limit as we’re approaching two from the right

of minus three and when i’m looking at this limit right here

so x is approaching two from the right and i have a non-zero

over something that’s going to zero the denominator is going to 0 right

2 minus 2. so i know that this limit is some infinity

again we have a non-zero that minus 3 is fixed it’s not changing

but the denominator is getting closer and closer and closer to zero

which means the whole expression is going larger and larger and larger

the question is is it plus or minus infinity

so now i’m going to check the sign now i’m approaching two from the right

for example two point one what happens at two point one

at two point one minus two two point one minus two is positive cube

00:42

root of a positive is positive so that’s that limit right there

positive infinity [Music] and just for fun what’s the limit as we

approach two from the left so can we approach two from the left um

for example what’s you know 1.9 1.9 minus 2 that’s a negative number can

we take the cube root of a negative number yes so 1.9 minus 2

that’ll be a negative and i do a cube root oh wait a second

um i just realized uh that’s not positive infinity i forgot to

take into account the numerator so first of all let’s look at both these limits

i have a non-zero over a zero so i can see both of these are sum infinity

now when i approach two from the right like 2.1

00:43

this is positive down here what’s a negative over a positive

is negative and here when i have neg 2 from the left like 1.9

then that’s going to be negative and what’s a negative over negative

is positive yeah so i almost missed that limit there by not paying attention so

the two-sided limit right here we would say this limit right here

the two-sided limit right here does not exist [Music]

so there’s an example of showing some local behavior

what’s happening around two really close to two and

we have a limit here that doesn’t exist because the behavior from the right is

minus infinity and the behavior from the left is positive infinity

so when we’re looking at two here and we’re looking from the left it’s

00:44

going to grow up and we’re looking from the right it’s going to fall down

okay so next example okay now we’re going to start talking about

vertical isotopes so those limits that we were just finding they were

with the goal of talking about vertical isotopes

so vertical isotopes is when you have unbounded growth

and we’re going to see the precise definition of what a um

limit is of what what a vertical isotope is

in other words when you were in calculus 1 you didn’t know about limits yet

and so you talked about vertical isotopes completely different

00:45

here we’re going to talk about vertical isotopes and we’re going to be much more

precise so vertical isotopes you need a limit first

in other words never say something is a vertical isotope

unless you have proof unless you have backup unless you have a reason

if you have a reason to say something is a vertical isotope

then you can say it so what are your reasoning well

if you have one of these limits if you have show one of these limits first

then you can claim you have a vertical isotope

now a vertical isotope of course is a line x equals c that’s a vertical line so

if you know one of these limits holds then x equals c is a vertical isotope

now the reason why you only need one of those limits is

because for some functions you may not even have

the possibility of some of those limits to make sense for example

00:46

maybe we have a function that looks like this we have a vertical isotope

and all right so that’s not a vertical isotope we need a vertical isotope

something like say something like this that’s a little bit more vertical

but i’m going down like this and so i have a vertical isotope right there

and maybe this function just goes off to wherever

but the point is is that this function may not even be defined over here

so if this is my vertical isotope x equals c

and we have the limit from the left at c is minus infinity so this growth right

here is going to minus infinity right here it’s just becoming unbounded below

we have this limit right here but the function may not be defined over here at

all so you may not even be able to take a limit from the left

so this uh yeah so this limit is from the right

00:47

so this limit from the right is minus infinity and you may not even be able to

talk about a limit from the left because

there may not be any points over here on the graph

it just may be empty over here so this right here would not exist does not exist

so you don’t need to talk about both sides to find a vertical isotope

so when i’m looking at this uh definition right here

all we need to know is at least one of those following limits

are true now you can have multiple ones true

you could have on this example right here i could have it from the left going

up or from the left going down you know i could have it going down and

right there or i could have it growing going unbounded right there so we you

could have multiple limits uh holding but you don’t have to

so the following example demonstrates in fact not all rational functions

have vertical isotopes so here’s an example of something that does not have a

00:48

vertical isotope so why not well if we try to find any of these limits

that we just talked about here those limits right there one two

three four five six none of those limits will work there’s no value for c

which that you can approach to get an infinite limit

there’s just no value so when we look at this uh

problem right here we’ll just simply say the function is continuous on its

domain because it’s a rational function and there’s no vertical isotopes so

if you try to find a c where the denominator is zero you won’t

be able to find one now contrast that with this one right here though

what if i change that to a minus ten right so this right here it’s positive ten

this one’s a minus ten now if we look at something like this

00:49

we can factor so we have f of x x to the third over x minus two times x

plus five so in this case we have vertical isotopes

because now notice i’m saying there’s vertical isotopes

but remember i just said you have to back it up with some limits

you can’t just say vertical isotopes and that’s it you have to say

why why is there a vertical isotope at x equals two in fact i’m saying

there’s two vertical isotopes so i got to back up each vertical

isotope with a limit so the limit i chose to find are these two right here

let’s make sure these limits hold so first off you have to figure out

what limits you want to look at right so let’s start this problem over again

and i’m looking at that function right there and i’m asking myself

what limits am i going to find i know that i need to find

some limits to find a vertical asymptote you can’t just say the denominator is

zero we just looked at some examples where that doesn’t work

00:50

so what are the places where i need to look so

i’m going to look at the function here f of x and i’m going to try to look where

the denominator is zero so we’re going to factor this this is what um

x minus two x plus five if we factor it [Music]

now if you were in pre-calculus you might jump to the conclusion

x equals two is vertical isotope it makes the denominator zero

x equals minus five that’s a vertical isotope it makes the denominator zero

but that’s just simply not true for example if i put x minus two up here [Music]

you could use the same reasoning x minus two

it makes the denominator zero it’s a vertical asymptote

minus five it makes a denominator zero it’s a vertical isotope

but it’s not a vertical isotope this one is

x equals two is not a vertical isotope in fact there’s a whole at

00:51

x minus two so anyways we have x to the third up here

we’re starting to think two and minus five are vertical isotopes

but before we can jump to this or if you want to say this you have to back it up

with some limits so i’m going to look at the limit as i approach minus

2 of this this function right here [Music]

and i’m going to look at the limit as i approach minus 5 right here

so i’m going to look at this limit right here [Music]

in order to say vertical isotope you have to find some limits

now in order to find these two limits here the approaching two from this one

and approaching the minus five from this one in order to find these limits

these are both two-sided limits so before i find the two-sided limits

i’m going to first go and find the one-sided limits

so first let me find the limit from the left

00:52

what is the limit from the left first so what is the limit for the left well

the numerator is going to an eight and the denominator is going to zero

so this is going to be some infinity right here

we’re going to a fixed number on top we’re going to an eight

and here the denominator is becoming unbounded

it’s uh going to zero and so the whole expression is becoming unbounded

now if we look two from the left like say 1.9

1.9 in here i’m going to get a negative and a positive

and so the whole expression will be negative and then

1.9 up here that’s still positive so this is going to minus infinity there

now if i look at from the right i’m going to approach 2 from the right now

00:53

x to the third x minus 2 and then x plus 5. now if i look at from the right i’m

approaching two from the right so now think of 2.1 for example 2.1 is

still going to close to an 8 at the top but the 2.1

down here in the denominator is positive now two point one minus two so that’s

positive so two point one plus five that’s positive

so positive times a positive underneath a positive so that’s positive infinity

so the two-sided limit actually didn’t exist now to say that i have a vertical

isotope x equals two all i really need is one of these two

limits here we don’t need both as soon as you have one of them you’re

already justified in claiming you have a vertical isotope

i found both of them just just to find them

now if you’re trying to sketch the graph of this then you would like to know

behavior on both sides of two because that could help you sketch the

00:54

graph but just in terms of finding the vertical isotopes you can

claim x equals two is a vertical isotope not because the denominator is zero but

because this limit is infinity minus infinity or you could

use this limit as positive infinity now let’s try to back up the claim that

minus 5 is a vertical isotope so minus 5 makes the denominator 0

and it doesn’t make the numerator equal to 0 so you’re thinking it has a

vertical isotope well let’s back that up with the limit

so i’m going to look at minus 5 of the function here the finest limit

now again this is a two-sided limit so i’m going to look at both sides

i’m going to look at limit from the left and i’m going to look at the limit from

the right so minus 5 from the right minus 5 from the right so x to the third

over x minus 2 times x plus five [Music] let’s find these two limits here now

00:55

notice both these limits the numerator is going to a to a number

so minus five to the third and then same thing with the from the right minus

five to the third but the denominator is going to zero so

minus five plus five that’s going to zero there

so um to find the limit from the left of minus five

i choose a number just a little bit to the left of minus five here

so here’s minus five so what’s a number like just a little bit to the left

like minus five point one right a little bit to the left

so minus five point one i’m going to get a negative on top i’ll just put a

negative and if i do 9 minus 5.1 minus 2 i get another negative

and minus 5.1 plus 5 that minus 0.1 is going to make that whole thing a

negative so i’m gonna get another negative

so negative over a positive is negative so this is negative infinity here

00:56

now we already have enough to claim this is true

x equals minus five has to be a vertical isotope

at least from one side from the left it’s a vert

to vertical isotope but just for the fun let’s go look from the right also

so if i approach minus five from the right

now let’s think of something like minus 4.9

right that’s a little bit to the right so minus 4.9

on top it’s going to be a negative i’ll put it over here um if i use a minus 4.9

up here it’s going to be a negative right so looking like negative and

what’s minus 4.9 minus 2 that’s also negative minus four point

nine minus two that’ll be negative what about minus four point nine plus

five that’ll be positive so positive over negative i’m sorry a

00:57

negative over a negative is positive so this limit didn’t exist

either from both sides at minus five we can write that

but don’t really need to we know from both sides as a vertical isotope one

side’s going down one side’s going up but any case the problem asks us to find

vertical isotopes we found both vertical isotopes and all

we really needed to show was these two limits right here

if we had found those two limits right there we could have backed up that

they’re both vertical isotopes just for gravy we also found those

limits for there [Music] all right so that’s determining vertical isotopes

and let’s see here [Music] all right so there’s f where we had x to

the third on top we had a minus ten when we had a positive ten here there

were no vertical isotopes because there’s no place where

the denominator is zero we have a minus 10 we were able to factor and we’re able

00:58

to get some isotopes um and so [Music] now i change the numerator again [Music]

so now let’s change the numerator to so the denominator is the same the

denominator still factors as x minus two x plus five

but now i’m getting a numerator and we can factor that numerator also

the factors as x x minus two [Music] and now we see the x y is twos will cancel

but now we have to recheck this here in fact let’s just recheck it all

so now the problem is not x to the third on top

it’s x times x minus 2 on top now you might say

look at 2 i get 0 down here and minus 5 i get

00:59

0 down here also so those are vertical isotopes

but remember we have to check them we have to check them with limits

so let’s do like we did before what is this limit right here

we approach two of f so it’s x x minus two x minus two x plus five

now before we check the limit from the left and the limit from the right

but this time i don’t think we need to because the x minus twos cancel [Music]

you could you can check them both from both sides if you wish

but you won’t need to so this will be two over seven

so as you approach two from both sides the height is approaching two over seven

now notice that two is not in the domain of the function

and what that means is that there’s a hole and the height is approaching

two sevenths when you get closer and closer to two

01:00

the height is approaching two sevens but there’s a hole there

not an isotope and then we can go and find this limit right here

as we found before we approach minus five from the left

of x over x minus two and then x minus two and then x plus five

now we can go cancel those x minus twos if we wish

we’ll just get x over x plus five and then now let’s try to find this

limit here so this one was not true [Music] this limit was not infinity so the

heights just approaching two over seven we have a hole there now for this one

i still cancel the x minus twos but now i’m trying to find this limit right here

now we’re approaching minus five from the left think of minus five right here

what’s the number to the left like minus five point one right [Music]

now the point is is that the numerator is going to minus five

01:01

and the denominator is going to zero so we’re going to a non-zero

over zero so that means we have infinity right here

the question is is it positive or minus infinity

so i’ll test minus five point minus five point one

that’ll make the numerator zero uh negative and minus five point one plus five 5

that’ll still be negative on bottom negative over negative is positive infinity

so we found that we found that limit right there it’s positive infinity

just think to yourself if i’m a little bit to the left of minus 5 like minus 5.1

what will be the sign of all that minus 5.1 will be negative

and this will be negative so negative over negative would be positive

so we got a we got infinity out for this limit

so that means this is a vertical isotope when we approach minus five from the

01:02

left the behavior is growing unbounded and so here’s an example where

the denominator was zero but you still have to back them up with limits

so not a vertical isotope this one is a vertical isotope x equals minus five

[Music] okay next example this example here

tangent x minus cotangent x here’s an example i want to show you an example of

something so we see an example where there were no vertical isotopes

we see an example where there was one there could be two

could you have infinitely many vertical isotopes and the answer is yes you can

look at a function like secant secant x or cosecant x or tangent x or

cotangent x or even tangent x minus cotangent x

this has a vert nuh uh infinitely many vertical isotopes

so the way you could look at that is by breaking it down and combining it all

01:03

together so it’s it’s a lot easier to look at something like

secant so if i were to look at something like secant or even cosecant

so remember the graph of sine looks like this [Music]

hits zero here and zero here and so if i was to go

sketch this sketch let’s let’s scale that down a little bit

so we can put the cosecant graph on here the skill that down here is

graphic sign here [Music] so wherever it hits zero right so cosecant is

one over sine right so now we’re worried about where sine is zero

01:04

so sine is zero right here and here and here

and it keeps hitting zero over and over again doesn’t it

and so the cosecant graph is right here is right here

we have a vertical isotope here we have a vertical isotope here

it keeps alternating back and forth so cosecant is this part these parts right

here in blue [Music] and we can see that wherever sign is zero there’s going to

be a vertical isotope so where sine zero right so here’s pi over two here’s pi

here’s three pi over two and here’s two two pi so it’s every pi

every pi sine is zero so every pi cosecant x has a vertical isotope so we

01:05

say something like pi k but also minus pi minus 2 pi and so on right

so the k can be take on positive and negative values

uh integers so when we look at something like this

we try to get tangent x to cot cotangent x combine it together into one

expression so we can try to find the vertical

isotopes so i switched to sines and cosines

and i combined it together and i used a trig identity

sine squared minus cosine squared is cosine 2x [Music]

so now we have one numerator and one denominator

we can ask where is the new where’s the denominator zero

and the numerator is not zero so the zeros of the sine cosine function

yield the vertical isotopes in other words where sine zero because

that that will make the denominator zero and where’s cosine zero and so we get

01:06

the values plus or minus pi over two we get that k because it’s going to

happen over and over and over again and we have to check for those values

the plus or minus pi over two plus k they don’t make the numerator zero

also and then we’ll know that those limits are either plus or minus infinity

and so we get infinitely many vertical isotopes and so there’s what a

graph of that would look like so this example shows you that it’s possible

to have infinitely many vertical isotopes

all right so let’s look at this example here evaluate this limit

so we’re going to take the same approach we did to the last example there

by finding a you know putting those fractions

together right so we have the limit as x approaches 0 of 1 over x minus 1

01:07

over x squared so to find this limit i’m not going to

try to find the limit of each piece because if you know if you just look at

the limit as one over x like we looked at that limit before

and we got different values from the left from the right

and when we looked at this limit right here we looked in fact we looked at both

of these limits um individually as each term here

in episode one but but you’re going to be talking about infinity minus some

infinity right so that’s not something that you’re going to be able to do

so i’m going to combine this together and say x over x that’s going to be x

minus 1 over x squared so i want to think about the function

in this form right it’s the same function but now i have one numerator and one

denominator so i’m going to be looking at the where the denominator is zero

01:08

which is x equals zero so thinking to myself

x equals zero is a vertical asymptote as i’m thinking to myself it now this

problem doesn’t mention anything about vertical isotopes but

i’m still thinking that x equals zero is a vertical system

but i i would need to justify it right you can’t just say that

i’m going to be looking at the limit from the left here

and the limit from the right now of course if you try to just substitute in zero

well this is a rational expression rational function it’s continuous on its

domain you can substitute in zero as long as

it’s in its domain but zero’s not in the domain

right so i’m going to be looking at the limit from the left and the limit from

the right [Music] if we try to substitute zero we’re going

to get minus one it’s going to minus one on top

it’s going to zero on the denominator so i said we’re going to get some infinity

or it’s not going to exist so to determine those i look at the

01:09

one-sided limits here so limit from the left again this is

going to minus one on top and zero on the denominator

but if i look at minus one from the left i can pinpoint

is it going to be positive or is it going to be minus infinity

same thing for this one the numerator is going to minus one

the denominator is going to zero so i know this is going to be some infinity

it’s going to a non-zero over zero so some infinity so the question is

positive or negative which one this could be both positives both negatives well

one could be the other all right so choose a number less than zero for

example negative point one negative point one minus one

that’s going to be negative in fact if you look if you think about this

the x is getting really really small really really close to zero

the minus one is going to dominate these are minuses on top

01:10

and in the denominator the x squares are going to dominate the x squares are

always positive this is always going to be a negative over a positive

so these are both negative infinities these are this is

minus infinity and in fact since we showed this limit here x equals zero is a

vertical asymptote even though it didn’t ask that as the

question it just said find this limit this limit is minus infinity

okay so next example well let’s look at a write-up real quick

so i’m going to rewrite the function with one numerator and one denominator

then i can make an argument these limits are both minus infinity

and there’s a little bit of a sketch of the graph right there you can kind of

see that right there that it falls it’s just falling down right there

01:11

it’s just falling down as you approach zero from both sides okay so next

take a break all right now it’s time for long-term behavior

so what we’ve been studying so far is long-term behavior

uh what we’ve been studying so far is short-term behavior

and so now it’s time to talk about long-term behavior

so here’s the definition of long-term behavior that we’re going to shoot for

so notice here that x is approaching infinity

and what that means is x is taking on values that are larger and larger and

larger so short term behavior x is approaching a c

a number so you could be approaching 2 for example you could be 1.9 1.999 1.999

but now x is approaching infinity what that means is that x can get as

01:12

large arbitrary large it can go to 10 x can be a thousand x could be a million

and so on so and then we have the same thing for

x is approaching minus infinity so now x is taking on values like

negative 10 negative 100 negative a million negative

a billion right and and also put the formal

definition there to give you a little peek what that means

we’re gonna have a separate episode for that for those people who

want to uh understand limits more precisely um for those calculus

students who want to do that all right so let’s look at those

a little bit more up close and so i made this argument earlier i want

to see if you can follow it now or hopefully you’re able to follow then

too but what does it mean to say x is approaching infinity and that the

limit is equal to l so i’ll sketch a graph here now we’re

01:13

doing long-term behavior long-term behavior is hard to model here

because you know i only have so much room on our screens but long-term behavior

means long-term behavior what’s happening out here no what’s happening out here

now as soon as you stop you’re not doing

long-term behavior so long-term behavior never stops what’s happening is you

never stop so here’s l it’s an out it’s along the y-axis what

does it mean to say that we approach l and it’s long-term behavior so [Music]

think of this as a horizontal isotope and long-term behavior x is approaching

infinity what that means is that the x’s are just

growing larger and larger and larger so this is this could be the behavior of

01:14

the function you’d go up and down up and down

but ultimately the long-term behavior is that it’s approaching

l eventually it has to get closer and closer and closer

and closer and closer to l it could it could

short-term behavior it could pass right through there

but the long-term behavior is that it approaches

l and in fact even even out here at one million

i still be able to cross over the boundary l but the long-term behavior

is that i’m approaching this l whatever the l is

l could be two you could have three you could have any height you want

but if you say that the limit is l that means the long term behavior

is l and the same thing for minus infinity the only difference is the behaviors

on the left side now right and that’s the two choices that we have

01:15

now if i look at the long term behavior way out here so x could take on values

like negative 10 negative 100 negative a million so we

have some l this time i’ll draw l down here

because l doesn’t have to be positive or negative but we have this boundary here

l but it’s not really a boundary in the sense that you can’t cross it

it’s the boundary in the sense that it’s the long-term behavior

so a function could come through here and pass through and go like this

but the long-term behavior is that it has to approach

l eventually it has to get closer and closer to

l and it can do that a variety of ways but it’s just long term so

you know think of stocks right you never want to think of stocks as short-term

behavior you never want to invest one day and

then take your money out the next day you want to think long-term behavior and

you know you often see the stock market going up and down up and down

01:16

and if you pay attention to the stock market

you know one day it could be really bad for you

you know two weeks later it might be really good for you but

what you’re really interested is long-term behavior supposedly all right so

now this is a different type of limit that we studied

then so far is because so far we’ve approached c

and now we’re approaching plus or minus infinity so

back to our limit laws that we had so we are assuming these two limits

exist as we approach infinity [Music] then we have the limit of a constant is

the constant the limit of a constant times the

function is the constant times the limit of the function

so these are all limit loss when we pre exo is approaching c

so we have the exact same limit laws but now we’re approaching infinity

in fact they’re also valid if you approach minus infinity

01:17

so i just wanted to mention that all these limit laws work exactly the same way

whether you’re not you’re approaching c or positive infinity or minus infinity

they all work the exactly the same way okay so now we have a

theorem just like we had for limits to infinity

now we have the same theorem or a similar theorem when we have

limits at infinity and what this is saying is that

if you have a real number a and a is a number on top and x is going to infinity

and you have x to the power that’s going to zero

so if your rational number like a fraction um as long as it’s defined for all x

even if you’re going to minus infinity then that limit is still

zero so here’s an example let’s work this example out here

01:18

so what is it x is going to infinity and let’s look at our function here

we’re looking at three x squared minus x minus two and we’re looking at five

x squared plus four x plus one and so what is this limit

okay so i’m going to try to use that theorem that we just talked about

and the way to do that is to divide everything through by the highest power

so this technique is called the highest power

so to find this limit here i’m looking at the numerator what’s the highest power

x squared and i’m looking at the denominator what’s the highest power

it’s also x squared so let me divide everything by x squared [Music]

so this will be five x squared over x squared

01:19

plus four x over x squared plus one over x squared

so that makes the problem look more complicated

but it actually is going to simplify the problem a great deal

so i just divided everything through by x squared

now what happens well some of it simplifies

what’s 3x squared over x squared right so that’s 3

and this will be minus x over x squared so that’ll be minus one over x

and this will be minus two over x squared and this will be a five right here and

this will be a four over x and this will be still this will stay 1

over x squared so now we can use that theorem that we just saw

remember that theorem that we just saw a second ago

and it said the limit as x goes to infinity of a over x to the r

01:20

is zero so this is a real number up here and this is a power of x and this limit

is zero and so this limit so this r here here it’s just a one

here the r is a two we have a number and a number so this is going to zero

that’s going to zero that’s going to zero that’s going to zero

so this limit is just three fifths so that’s a really nice theorem that we

just had let’s scroll back to that this is a nice theorem here those limits

are zero there in one and two we have those two limits there these are

zeros right here that’s a zero and that’s a zero and they make solving

a limit like this much much easier because we can divide by the highest

power and the reason why we divide by the highest power

01:21

is to guarantee that we have a number a on top so when we look back at this

example right here i i looked at the highest power on top was an x square

so if i divide everything through by x squared it’s going to match here

but because the x square was the highest power these are going to only have

x’s on the bottom these are only going to have x’s on the bottom

same thing with this one right here what’s the highest power is x squared

so if i divide everything through by x squared they match here so i get a 5

and then now all the rest of them have powers of x on the denominator

so those go to zero zero zero and zero so all we’re left with is three fifths so

whatever this function looks like it’s its long-term behavior is three-fifths

so we’re going to evaluate the numerator and denominator

or divide divide them both by the highest power of x

01:22

and so the limit is just three-fifths all right so

here’s an example here where that technique doesn’t work

straight away you have to first do a intermediate step now a common mistake

the reason why i want to show you this example is

because the common mistake is to say infinity minus

infinity right the the limit as x goes to infinity of

x that first term that’s just infinity and

the limit of the second part is infinity also

so you’re tempted to say that’s infinity minus and then that’s infinity and

that’s zero and well that’s just simply not true so

we’re going to um first combine this together

01:23

and then we can see what the limit’s going to be so i’m going to say

if i can this will be x minus square root of x plus one and i’m going

to use an x plus square root of x plus one over

x plus square root of x squared plus one so i’m multiplying by one

and i’m using the conjugate so remember we use the conjugate to find limits

in episode one the difference is now is that instead of approaching a fixed

number we’re approaching infinity so we’re studying long-term behavior but

algebraically we use the conjugate which is similar to what we did before

and the reason why we use the conjugate is even though the denomina

the denominator doesn’t really change into anything nice

the numerator does x times x and then we’re going to get square root

of x squared plus 1 times positive x squared of x plus 1 so that’ll be x

01:24

squared plus 1 with parentheses and so now we can cancel the

um x squares x squared minus x square and so we end up with a minus one over x

plus x squared plus one now there’s two ways to find this limit here we can use

common sense or we can use highest power common sense to me would be okay you

know i i like common sense what do i mean by common sense well

x is going to infinity right it’s getting larger and larger and larger

like 10 100 200 a million what is this doing the numerator is

staying at minus one but the denominator is getting larger

and larger and larger right because it’s all

positives it’s got a squared it’s got a square root it’s all positive down here

01:25

so for example when x is 10 that’s going to be 10 plus more

when x is a hundred it’s going to be 100 plus more when x is a million

then it’s going to be a million plus more

right there’s no subtraction here you’re just doing a million plus

more right so as x is getting larger and larger this thing is getting larger and

larger and larger and larger so what’s happening to the whole expression

is the numerator gets larger and larger and larger and the new

sorry as the denominator is getting larger and larger and larger with the

numerator staying fixed so this whole thing is getting

zero it’s going to zero so that limits going to zero now maybe you

didn’t see that and wanted to understand the highest power and that’s fine

01:26

because when i say it’s common sense i mean like after you’ve done

a thousand examples any case let’s use highest power

so i’m going to divide by everything by x x is the highest power so x over x

and then square root of x squared plus 1 over x

now why is x the highest power well clearly if i look in the numerator there

is no power of x so i only have to divide the top by x

because i’m dividing everything through by x and i have to be

we have to maintain equality here so i don’t get that x from looking at the

numerator i get the x from looking at the denominator

x well x is to the first power now when i have square root of x squared

square root of x squared that behaves like an x so the highest power

is an x now when we look at this we can simplify this a little bit

01:27

minus one over x and we get one plus and here we’re going to get so the x

here is not inside the square root so i can write as x squared plus 1

and then bring that into the square root so that’s an x so when i divide

all of this by an x i just have all of this divided by x

but think of that x as square root of x squared so if i want to bring that

inside of the radical i’m going to say x squared

so this now goes for the whole numerator and denominator x squared plus 1 over x

squared so the point is is that now we can write it like this

1 plus right so we have minus 1 over x minus 1 over x 1 plus 1 plus

01:28

now we have square root of 1 plus 1 over x squared let’s put it like that

1 plus 1 over x squared all right so now we’re ready to use our theorem

this is a power of x over a under a number minus 1 and

so this is going to zero that’s going to zero

so we get zero over one plus square root of one

we get zero over one or zero over two we get zero [Music]

so this limit right here is zero and um to be honest right here i think

because everything’s positive and there’s really no question

that the denominator is growing larger and larger and larger

that should be zero but if you wanted to just check your work or just to

understand it better i divided everything by the highest power which was

01:29

x and then i you know changed that to square root of x squared and then

simplified it all and then we found the finally found the limit

the limit of this top part is going to zero the limit of this part right here

is going to zero and so this will be one plus square root of one plus zero

[Music] okay so there’s that limit there using the conjugate

okay so now let’s talk about horizontal isotopes

all right so there’s the write-up of that limit right there

and i i went ahead and divided by the highest power there so you can see that

all right next uh next part [Music] all right horizontal isotopes

01:30

and so i just want to remind everyone here that

in calculus we um don’t use the symbol minus infinity as a number we use it as a

as a process as a limit and it’s used to describe unrestricted growth

whether that growth is positive or negative but it’s unrestricted growth now

horizontal isotope so we’re going to approach infinity x is

going to approach infinity or minus infinity and if you have either one of those

limits then you know the line y equals l is called a horizontal isotope

so to find a horizontal isotope you don’t need both of those limits you just

need one however if you’re going to find all the horizontal isotopes

you’ll need to check both limits in that case

so here it says find the horizontal isotope of the graph of the function

01:31

so we need to find the limit as we approach infinity and if we’re going to

find all of the horizontal isotopes then we’ll need to find the

limit as we go to minus infinity so let’s look at this function right here

let’s approach positive infinity and that’s positive infinity and we’re

looking at 1 minus x over times 2 plus x all over 1 plus 2x

and then two minus three x [Music] now the way that we found limits as

we’re going to infinity before is we divided by the highest power so

in order to do that i’m going to multiply this all out

here so that doesn’t take very long but just expanding that out so we’re

going to get minus x squared minus x plus 2

and in the denominator we’re going to get minus 6x squared plus

01:32

x plus 2. so again i just multiplied all that out and simplified it

and then i’m going to multiply 1 times 2 1 times minus 3x

and then 2x times 2 and then 2x times -3 x and then just simplify it now

we can do some calculus here we can divide by the highest power

right so the highest power in the numerators is squared

the highest power and the denominators are squared so i’m going to divide

everything through by an x squared so divide by x squared plus

x over x squared plus 2 over x squared all right so now

we can simplify that would be the next step um this right here is going to be a

01:33

minus one and this right here is going to be a

minus six and the rest of these are going to go to zero

and so our answer here is going to be 1 6. so now we can say that y equals 1

6 is a horizontal isotope so i’ll abbreviate horizontal isotope by h a

y equals 1 6 is a horizontal isotope so we found the horizontal isotope there

now you might be wondering what about minus infinity

what if we approach minus infinity well we can do that actually pretty quickly

right here let’s put a minus underneath there and

let’s take the limit as we approach minus infinity at the same time

if we can let’s check so i’ll put a minus infinity there

01:34

what did we do from this step to this step all we did was expand that out that

had nothing to do with the limit so that equality is true now what do we

do from this step to this step well i divided everything through by the

highest power x squared in other words i just took the same

thing and wrote it under x wrote it over x squared so that step

also has nothing to do with the limit so this equality is true because of

algebraic manipulation so in other words this this equal sign

and this equal sign still hold whether you have plus or minus infinity

now it’s just this equal sign that might make a difference

now if we’re approaching minus infinity this is still going to zero and that’s

still going to zero and that’s still going to zero and

that’s still going to zero what about these well this is x squared

over x squared that’s still going to be minus one and

this is x squared over x squared so that’s still going to be minus six

01:35

so i’m still going to have one sixth so that’s still my only

horizontal isotope so in other words this line is a horizontal isotope

from both sides long term to the right and long term to the left they’re both

both sides are going to approach that isotope there all right so let’s look at

another example so i wrote it up with positive and minus infinity there

now not always can you take positive infinity and the limit as we approach minus

infinity at the same time so i’ll show you an example of this um now actually

that should say obviously this is one over six here

if my arithmetic is correct that minus one six

01:36

minus one over minus six is one sixth there all right so next example

and let’s look at this function right here now i see another uh problem here

it says find the horizontal isotopes so it should say isotopes with a

parenthesis with an s there because there could be more than

one so i should say find the horizontal isotopes

of the graph so we want to be able to find them all

so let’s do that now find the horizontal isotopes

so we’re going to look at the limit as we approach

positive infinity of this function right here

01:37

as we approach positive infinity of square root of two x squared plus one over

three x minus five and we want to look at the long-term behavior on the left

side also so we’re going to find the limit as we approach

minus infinity also so i want to keep that in mind as we’re taking this limit

because maybe we’ll be able to take both limits at the same time maybe we won’t

we’ll just have to see what we do so first step here

is i’m going to try to divide by the highest power

whenever we’re taking limits as whenever we’re studying long-term behavior

we’re approaching plus or minus infinity a common thing to try to do is to

use divide by the highest power the denominator the highest power is

clear it’s just an x the numerator well we have an x squared

but we have a square root around it so the highest so that’s behaving like an x

so the highest power here is going to be an x so i’m going to divide everything

01:38

by an x so divide all of this by an x and i divide the

numerator by an x or sorry the denominator by x

all right so good so just divide it the numerator by x and the denominator by x

now we can break this up a little bit by the way that would also work if you

put minus infinity there i’m not going to do it yet but dividing both sides by x

is okay now i’m going to um in the in the numerator i’m going to say this is

two x square plus 1 and then i want to pull that x inside of the

square root but i want to be a little bit careful here

i’m going to say this is square root of x squared

here and then this is on the denominator i’m going to say this is 3x over

01:39

x and minus 5 over x so i change this x to a square root of x squared

now is that always true well the answer is yes as long as x is positive

for example three is equal to square root of three squared right

because that’s just the square root of nine but is minus three equal to minus 3

squared square root so if i plug in a 3 here into this

if i plug in 3 into here yeah it is true but if i plug in a minus three here and

a minus three here then it’s not true is it because this is

still square root of nine this is still three

this one’s not true so in other words when i look here at this x and i write

it as the square root of x squared it is true because the x’s are

approaching positive infinity and what that means is the x’s are

01:40

positive think of x is like a 10 a 3 a 3 million so i can always write

x as the square root of x squared so the point is is that our next step

looks like this so our next step is say equals we have the limit

as x goes to positive infinity in the denominator we just have a 3 over

3 minus 5 over x all right that’s the denominator with

the numerator i have a square root divided by square root so i can

write it like this square root of two x squared over

x squared plus one over x squared and the x squares here will cancel

so the limit as we approach positive infinity of

01:41

square root of two plus one over x squared

over three minus five three minus five over x

and so now we’re ready for our theorem we can use our theorem now

as x’s go to infinity we can say this limit right here is going to zero

and that limits going to zero and so we’re gonna get square root of two over

three and that’s our limit here so the limit as we approach positive infinity

the square root of all this is square root of two over three

that’s the long term behavior if you look at the graph of this function right

here and you look at it long term it’s approaching this height right here

now in the next episode we’re going to talk more a lot uh we’re going to cover

is curve sketching and so definitely horizontal and vertical isotopes

will have a big impact on our on our curved sketching knowing the long

01:42

term behavior of a function is very important now what about if i

try to go minus infinity this this step will be the same but this

step right here will not be the same if we’re going to go to minus infinity

then this will not be equals here so let’s look at that limit there

let’s erase all this here and let’s erase this here

and put minus infinity and minus infinity

now from here to here what did we do we just took the numerator and divided

divided by x and took the denominator and divided by x

and that had nothing to do with what we’re approaching

now what about from here to here now remember the x is not always equal to

square root of x squared if we’re approaching minus

01:43

infinity that means this x here is negative for example negative 10

negative 100 negative a thousand that x’s are always negative don’t don’t

think just because you see an x that’s positive that x’s could be negative

so i have to put a negative in front of here when this is going to minus

infinity so now we can work this out very similar

to what we just did we can pull the square roots together

so i’m going to say minus square root of 2 over

x squared sorry 2x squared over x squared and then plus 1 over the x squared

and then still we have 3 minus five over x

so the only difference than what we had before was this minus sign here

so we still have the same denominator there the x’s cancel

minus five over x so this is they both have square root so i can

write them together inside the square root and this is 2x squared over x squared

01:44

this is one over x squared but the key difference is that minus is

outside of the square roots so now this is going to zero this is going to zero

that’s going to two so now we’re going to get negative

when we pass the limit we’re gonna get negative square root of two

over three and so what’s happening here is that this function on the long term

behavior on the left side we’re approaching and so we have a different

horizontal asymptote so y equals negative square root of 2 over 3 is also

a horizontal isotope so this is an example of something that has

two horizontal asymptotes now hopefully that’s not too surprising to you

remember the function arc tangent remember our tangent it just looks like this

01:45

has a horizontal tangent has a horizontal isotope

and this is y equals pi over two and this is y equals minus pi over two

and arc tangent goes like that and over here goes like this and so that has the

tangent inverse function has two horizontal isotopes also

um but so does this function right here it doesn’t have to be an inverse tree

this given example right here um you know that has also has two horizontal

isotopes all right so there’s our right up dividing both

numerator and denominator by x and i’m going to use the property of limits

and now when i approach a minus infinity i have a minus in there

01:46

and so we see that that’s minus square root of two over three

all right so we have two horizontal isotopes all right so let’s look at one last

example the example we just did had a square root of an x squared in it

so now what happens if we have a fourth root of an x to the fourth

so let’s find these limits here i’m going to approach infinity from the right

our positive infinity so we have x and then the fourth root of

x to the fourth plus one and so now i’m going to divide by the highest power

of x in the numerator the highest powers and x

a fourth root of an x to the fourth that behaves like an

x so the highest power is going to be an x so i’m going to divide everything

01:47

through by x x over x and we have fourth root of x to the fourth plus one

over x also i should like to group them to make it a little bit easier to read

but so the numerator is easy x over x is just one

i want to start down here actually so here we go we have positive infinity

we have one over now i cannot just bring this x inside the fourth root

so i’m going to say this is so fourth root of x to the fourth plus

one but for this x i’m going to write it as a fourth root of an x to the fourth

so that’s what x is but that’s equal to x only if we’re approaching

01:48

positive infinity if we’re approaching minus infinity then

i would need a minus sign right here okay so now we can simplify a little bit

further we have one over and we have the fourth root of all of this

and so that’s going to be x to the fourth over x to the fourth that’ll be one

plus and this will be one over x to the fourth and so this will be one over

the fourth root of that’s going to zero it’s a fixed number and that’s going to

infinity so that’s going to zero so this will be the fourth root of one

which is just one so the limit here is one so y equals one is a

horizontal isotope so we took the limit as x went to infinity

and we got out one so what y equals one is the horizontal isotope

01:49

now we want to compute the limit as we go

from minus infinity so what will change what will change if we do minus infinity

now right same function minus infinity now so now we’re looking at long-term

behavior on the left side so this will be minus infinity now

now what did we do from here to here all we did is divide through by the x

so that’s the same whether or not this is plus or minus infinity x over x

is the same so that is a good equal sign there now what about here

well what did we do x over x is still one regardless if you’re going to

minus infinity or not so x over x is still one what about this though

i’m switching the x to fourth root of x to the fourth

are these always equal to each other and the answer is no

if x is negative well then that’s a negative

01:50

but it but if x is negative here that’s the fourth power that will

that will remove the negative and this whole thing will be positive so we’re

going to put a negative in front of here if we’re going to go to negative

infinity there needs to be a negative here so now this limit right here if i

simplify all of this i’ll just have a negative out front

so that negative is out of the fourth roots

so when i look at the fourth root of all of this the negative is still out there

so this will be now be negative one i still have one over

and this will be negative fourth root of one so that’s still negative one so now

we have another horizontal isotope y equals negative one is also

a horizontal isotope so here’s an example of another problem

with two horizontal isotopes okay so again i compute the limit from the

01:51

the long-term behavior on the right-hand side the limit as we approach positive

infinity now some people don’t write the positive infinity symbol

and so i didn’t in that case but i get one for that limit

and then you have to remember the fourth root of x to the fourth is minus

x when x is negative right so we need a negative sign there

and so then we’re able to find the limit as we approach minus

infinity so in this case we get two horizontal isotopes

all right so um now let’s look at some exercises

okay so let’s look at some exercises here [Music] and here we go some exercises

and so exercise one is looking at short-term behavior

so we’re approaching a number x is approaching a number so that’s

01:52

short-term behavior for example number one is a non-zero

over a zero so you can tell that that is short-term behavior there

and and we’re looking at some limit there going to infinity um

same thing on number two you’re given a function

and you’re asked to look at some limits and so for example number one you take

the limit as x approaches two from the right of that function right there

and then all the way to number four okay so those are some exercises like the

beginning part of this episode and then here are some exercises that look like

the latter part in the sense that we have to find

all of the isotopes so yeah so now when someone says find the isotopes

you have to look for vertical isotopes and horizontal isotopes you have to put

it all together so exercises four there um

01:53

so we did some exercises like that in a previous in previous episodes where

we’re given some information that describes the graph

and then you have to go sketch the graph so that those are really fun problems i

like those problems they help you make sure that you understand what’s going on

and so here we have some more algebra to find some limits

and then number six we have some trigonometry to find some limits

so make sure you have the graphs of sines and cosines and

secants and cosecants and make sure you kind of have those

in your mind so you can get help with those limits

okay and so then there’s some last limits there exercise seven is the last one

where we’re looking at long-term behavior so the x’s are approaching some

infinity and we want to look for the long-term behavior of those functions

and so there’s those exercises there um so i want to say that um

01:54

if you have a question about any of those exercises if you try them and you

get stuck leave a uh comment below and i’ll make a video about those and

you know let me know what you think about those exercises the next video is

inflection points and concavity and we’re going to cover a lot of curve

sketching we’ll also cover vertical tangents and vertical cusps

so i look i look forward to seeing you on that video

and uh i’ll see you next time if you like this video please press this button

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