Finding Horizontal Asymptotes and Vertical Asymptotes (Limits Involving Infinity)

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

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

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

00:01
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

00:02
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

00:03
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

00:04
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

00:05
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

00:06
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

00:07
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

00:08
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

00:09
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

00:10
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

00:11
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

00:12
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

00:13
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

00:14
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

00:15
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

00:16
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

00:17
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

00:19
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

00:20
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

00:21
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

00:22
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

00:31
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
and subscribe to my channel now i want to turn it over to you
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About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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