### Video Transcript

In this video, weβre going to learn
how to use limits to help us understand the asymptotic behaviour of functions. Weβll recall what it actually means
for a function to have horizontal and vertical asymptotes and use the laws of limits
to help us find the location of any asymptotes.

We begin by recalling the
definition of a horizontal or vertical asymptote. They are, respectively, horizontal
and vertical lines such that the distance between the curve and the line approaches
zero as either the π₯- or π¦-coordinate approaches β. The graph of π¦ equals one over π₯
squared, for example, has a horizontal asymptote given by the equation π¦ equals
zero and a vertical asymptote given by the equation π₯ equals zero.

Itβs important to realise that a
function can have absolutely zero horizontal or vertical asymptotes. For example, π¦ equals sin of π₯
and indeed π¦ equals cos of π₯ have no asymptotes at any point. Similarly, the graphs of
polynomials are smooth and continuous, with no asymptotes of any kind. What weβre looking to do is
formalise this definition using limits. If we let π be defined on some
open interval π to β, then if the limit as π₯ approaches β of the function is equal
to some constant value πΏ, this means that the value of the function can be made
arbitrarily close to πΏ by requiring π₯ to be sufficiently large. Visually, that could look a little
something like this.

Alternatively, we can say that, for
a function to find on some open interval negative β to π, then if the limit as π₯
approaches negative β of the function is equal to some constant πΏ. Then the values of the function can
be made arbitrarily close to πΏ, this time requiring π₯ to be sufficiently large
negative. And once again, we have a diagram
representing that scenario. Notice that each time, the graph
approaches the horizontal line π¦ equals πΏ as either π₯ gets sufficiently large or
sufficiently large negative. And that helps us with the formal
definition for a horizontal asymptote. We say that the line given by the
equation π¦ equals πΏ is a horizontal asymptote of the curve π¦ equals π of π₯ if
either the limit as π₯ approaches β of π of π₯ equals πΏ or the limit as π₯
approaches negative β of π of π₯ equals πΏ.

And what about our vertical
asymptotes? This time, we say that the line π₯
equals π is a vertical asymptote of π¦ equals π of π₯ if at least one of the
following is true. We can look at one-sided
limits. If the limit as π₯ approaches π
from the left of π of π₯ is either positive or negative β. If the limit as π₯ approaches π
from the right of π of π₯ is either positive or negative β. Or simply, if the limit as π₯
approaches π itself of π of π₯ is either positive or negative β. And this is for a function to find
on some open interval that contains π except possibly at π₯ equals π. In particular, with rational
functions, those written as quotients of two functions, vertical asymptotes are
vertical lines which corresponds to these zeros of the denominator of our rational
function. Weβre now going to have a look at
how to apply these definitions.

Find the horizontal and vertical
asymptotes of π of π₯ equals two π₯ cubed plus two over π₯ cubed minus one.

Letβs begin by recalling the
definition for a horizontal and vertical asymptote. We say that the line π¦ equals πΏ
is called a horizontal asymptote of the curve π¦ equals π of π₯ if either the limit
as π₯ approaches β of π of π₯ is equal to πΏ or the limit as π₯ approaches negative
β of π of π₯ is equal to πΏ. And we say that the vertical line
π₯ equals π is called a vertical asymptote of the curve π¦ equals π of π₯ if at
least one of the statements is true. More generally, with rational
functions, those written as the quotients of two functions, vertical asymptotes are
the vertical lines which correspond to the zeros of the denominator of the
function.

So in this question, π of π₯ is
equal to two π₯ cubed plus two over π₯ cubed minus one. So weβll begin by finding the limit
as π₯ approaches β of two π₯ cubed plus two over π₯ cubed minus one. We usually begin by checking
whether we can do this with direct substitution. However, if we were to use direct
substitution, weβd end up with β over β, which we know to be an indeterminate
form. So instead, weβre going to look to
manipulate our expression somewhat. Weβre going to divide both the
numerator and the denominator of our quotient by the highest power of π₯ in the
denominator. Thatβs π₯ cubed. That gives us two π₯ cubed over π₯
cubed plus two over π₯ cubed all over π₯ cubed over π₯ cubed minus one over π₯
cubed.

Well, of course, two π₯ cubed over
two π₯ cubed is two. And π₯ cubed over π₯ cubed is
one. And now, we see that as we let π₯
approach β, two over π₯ cubed and one over π₯ cubed each approach zero. And so the limit as π₯ approaches β
of our function is two over one, which is simply two. And in fact, it follows that as π₯
approaches negative β, we get the same result. So the line with equation π¦ equals
two is a horizontal asymptote of our function.

We now move on to finding the
vertical asymptote. Weβre looking to find the value of
π₯ such that the limit as π₯ approaches this value of the function is either
positive or negative β. And of course, the way we achieve
this with rational functions is we look for the zeros of the denominator. Well, the denominator of our
function is π₯ cubed minus one. So we need to find where π₯ cubed
minus one is equal to zero. Letβs add one to both sides of our
equation and such that π₯ cubed is equal to one. And then, weβll take the cube root
of both sides. And we find that the real solution
to the equation π₯ cubed minus one equals zero is π₯ equals one. And we found the vertical asymptote
of our function. Itβs π₯ equals one.

In our next example, weβll look at
a function that requires a little more manipulation.

Find the horizontal and vertical
asymptotes of π of π₯ equals one over π₯ minus one minus two.

We begin by recalling the
definition of a horizontal and vertical asymptote. We say that the line π¦ equals πΏ
is a horizontal asymptote of a function π¦ equals π of π₯ if the limit as π₯
approaches either negative or positive β of π of π₯ is equal to πΏ. Similarly, π₯ equals π is a
vertical asymptote of a function π¦ equals π of π₯ if any of the six statements are
true. That is, the limit as π₯ approaches
π from the left of the function is either positive or negative β. The limit as π₯ approaches π from
the right of the function is either positive or negative β. Or the limit as π₯ approaches π of
the function is positive or negative β.

Now, our function π of π₯ is equal
to one over π₯ minus one minus two. So weβll begin by evaluating the
limit as π₯ approaches β of one over π₯ minus one minus two. And in fact, very little
manipulation is required here. As π₯ gets sufficiently large, one
over π₯ minus one approaches zero. And so the limit as π₯ approaches β
of one over π₯ minus one minus two is zero minus two, which is simply negative
two. And in fact, as π₯ approaches
negative β, one over π₯ minus one still approaches zero. So the limit as π₯ approaches
negative β of π of π₯ is equal to negative two. And we find the equation π¦ equals
negative two is a horizontal asymptote of our function.

Next, weβll consider the vertical
asymptotes. Less formally, we say that, with
rational functions, thatβs those written as the quotient of two functions, vertical
asymptotes are the lines which correspond to the zeros of the denominator of the
function. So in fact, weβre going to begin by
subtracting two from one over π₯ minus one by creating a common denominator. We multiply two over one by π₯
minus one over π₯ minus one. And then, we can subtract the
numerator. So we have one minus two times π₯
minus one over π₯ minus one. We simplify. And we find that π of π₯ can be
written as three minus two π₯ over π₯ minus one.

Weβre interested in the values of
π₯ so that our denominator π₯ minus one is equal to zero. Well, if we solve this equation for
π₯ by adding one to both sides, we find that π₯ is equal to one. And so we found the horizontal and
vertical asymptotes of π of π₯ equals one over π₯ minus one minus two. The horizontal asymptote is π¦
equals negative two. And the vertical asymptote is π₯
equals one.

Weβll now consider how we might
find asymptotes for a nonrational function.

Find the horizontal and vertical
asymptotes of the function π of π₯ equals three π₯ minus sin π₯.

We recall the horizontal line π¦
equals πΏ is an asymptote to the function π¦ equals π of π₯ if the limit as π₯
approaches either positive or negative β of π of π₯ is equal to πΏ. Then, for a vertical asymptote, we
say π₯ equals π is an asymptote if the limit as π₯ approaches π from the left of
π of π₯ is either positive or negative β. If the limit as π₯ approaches π
from the right of π of π₯ is either positive or negative β. Or if the limit as π₯ approaches π
of π of π₯ is either positive or negative β.

Now, in this question, π of π₯ is
given by three π₯ minus sin π₯. So letβs begin by finding the limit
as π₯ approaches β of three π₯ minus sin π₯. Then, we recall that the limit of
the sum or difference of two functions is equal to the sum or difference of the
limits of those respective functions. And we can write this as the limit
as π₯ approaches β of three π₯ minus the limit as π₯ approaches β of sin π₯. As π₯ approaches β, three π₯ itself
also approaches β whereas sin π₯ can only take values in the closed interval from
negative one to one. And so the limit as π₯ approaches β
of three π₯ minus sin π₯ is β. This isnβt a constant value as
required of πΏ. So thereβs no horizontal asymptote
here.

We should check the limit as π₯
approaches negative β. Once again, we split it up into the
limit of three π₯ minus the limit of sin π₯. As π₯ approaches negative β, three
π₯ itself also approaches negative β. Once again though, sin π₯
oscillates between negative one and one. This has very little impact on a
number as large as negative β. So the limit as π₯ approaches
negative β of three π₯ minus sin π₯ is negative β. And we can say that there are no
horizontal asymptotes.

Now, weβll consider the vertical
asymptotes. Weβll split our limit up. And weβll look for the limit as π₯
approaches π of three π₯ minus the limit as π₯ approaches π of sin π₯. We want to find a situation where
this might be equal to either positive or negative β. Well, we saw that the only way for
the limit of three π₯ to be β is if π₯ itself approaches β. Similarly, for three π₯ to approach
negative β, π₯ itself must approach negative β whereas the range of sin π₯ is the
closed interval from negative one to one. So the only way for the limit of
our function to approach positive or negative β is if π₯ itself approaches positive
or negative β. So actually, there are no vertical
asymptotes either. And so the function π of π₯ equals
three π₯ minus sin π₯ has no horizontal or vertical asymptotes.

By considering the three examples
weβve looked at so far in this video, you might be inclined to believe that any
function that is not a rational function will not have horizontal or vertical
asymptotes. In fact, this is not true. And weβre able to quote a couple of
these. We know that the limit as π₯
approaches zero from the right of the function the natural log of π₯ is negative
β. And so π₯ equals zero is a vertical
asymptote for the function π¦ equals the natural log of π₯. And in fact, the same is true for
π¦ equals log base π of π₯ provided that the value of π is greater than one.

In this video, weβve learned that
the line π¦ equals πΏ is called a horizontal asymptote of the curve π¦ equals π of
π₯ if either the limit as π₯ approaches β of π of π₯ is equal to πΏ or the limit as
π₯ approaches negative β of π of π₯ is equal to πΏ. Similarly, we found that the
vertical line π₯ equals π is a vertical asymptote of the curve π¦ equals π of π₯
if any of the six statements shown are true. We saw that, less formally, with
rational functions, vertical asymptotes correspond to the zeros of the denominator
of that function.