We will consider the function from last week:
√
x
f (x) :=
.
x+3
Notice the behavior of this function as x gets large. Although the function f (x)
is decreasing, it seems to be decreasing toward a finite value.
How can we find the value? We need to consider the limit as x → ∞.
√
1
x
= lim 1/2
lim
x→∞ x
x→∞ x + 3
+ 3x−1/2
=0
The limit is seen to be zero, because x1/2 gets large as x approaches infinity,
and x−1/2 goes to zero as x approaches infinity.
Notice the behavior in the graph of f (x): as x gets large, the graph gets
closer to the line y = 0. We say that the line y = 0 is a horizontal asymptote of
f (x).
To find the horizontal asymptotes of a function, we need to take the limits
as x → ∞ and as x → −∞. These limits might be different; so a function can
have up to two horizontal asymptotes.
Definition 0.1. A function f (x) has a horizontal asymptote at y = a if
limx→∞ f (x) = a or limx→−∞ f (x) = −a.
This is one type of “infinite limit”- a limit as x approaches ∞ or −∞.
However, there is another type of “infinite limit” that we are interested in.
Consider the function
1
.
g(x) =
(x − 5)
This function is not defined at x = 5. What happens for values of x that are
close to 5? If x is a number that is slightly smaller than 5, then g(x) will be
a very negative number. In fact, g(x) can be made as negative as you want by
taking x close to 5.We say
lim g(x) = −∞.
x→5−
Similarly, if we plug in a value that is very slightly larger than 5, for x, g(x) is
a very large positive number. In fact g(x) can be made as large as you want by
taking x close enough to 5. We say
lim g(x) = ∞.
x→5+
Either one of these conditions, on its own, is sufficient for x = 5 to be a vertical
asymptote of g(x).
Definition 0.2. A function g(x) has a vertical asymptote at x = c if at least
one of the following four things happens:
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• limx→c− g(x) = ∞
• limx→c− g(x) = −∞
• limx→c+ g(x) = ∞
• limx→c− g(x) = −∞
Note that we only need one of these things to happen. It may happen that
one of the left-hand and right-hand limits is finite, and the other one is infinite.
Notice that this definition does NOT say anything about whether g(c) is
defined. It is entirely possible for g(x) to be defined at x = c. In other words,
a function may be defined at an x-value at which it has a vertical asymptote.
Here’s an example:
Consider the rational function
f (x) =
x2 + 3x + 2
.
x2 + 5x + 6
Find the horizontal and vertical asymptotes of this function.
We will start by finding the horizontal asymptotes:
x2 + 3x + 2
x→∞ x2 + 4x + 4
1 + x3 + x22
= lim
x→∞ 1 + 4 + 42
x
x
1+0+0
= lim
x→∞ 1 + 0 + 0
=1
lim
We can immediately see that limx→−∞ f (x) = 1. So there is only one
horizontal asymptote: the line y = 1.
We know that this function is continuous and finite for all x for which the
denominator is nonzero. So the first step for finding the vertical asymptotes of
a rational function is to look for the points at which the rational function is
undefined; that is, where the denominator is equal to zero. We notice that the
denominator is x2 + 4x + 4, which factors into (x + 2)(x + 3). So the only places
we could possibly have a vertical asymptote are at x = −2 and x = −3.
To check whether these are vertical asymptotes, we need to consider the
limit of f (x) as x approaches −2 and as x approaches −3. Let’s start with
x2 + 3x + 2
x→−2 x2 + 5x + 6
(x + 2)(x + 1)
= lim
x→−2 (x + 3)(x + 2)
x+1
= lim
x→−2 x + 3
= −1
lim f (x) = lim
x→−2
2
In particular, this is finite, so we do not have a vertical asymptote at x = −2.
This is an important point: an asymptote is not just an x-value that is not
in the domain. An asymptote is an x value where the function f has an infinite
limit. That’s not the same thing.
Now, let’s check the limit as x → −3:
lim f (x) = lim
x→−3
x→−3
x+1
x+3
Now, the denominator goes to zero but the numerator doesn’t. As we approach
−3 from the left, f (x) approaches −∞. As we approach −3 from the right, f (x)
approaches ∞. So f (x) has x = −3 as its only vertical asymptote.
We have a “hole” at (−2, −1). These “holes” are often referred to as removable discontinuities.
Let’s consider another example:
Find all the vertical and horizontal asymptotes of
e2x
.
−1
ex
First, let’s look for the horizontal asymptotes. The limit as x approaches infinity
of f (x) is
e2x
x→∞ ex − 1
e2 x · e−x
= lim x
x→∞ (e − 1) · e−x
ex
= lim
x→∞ 1 − e−x
=∞
lim f (x) = lim
x→∞
So f (x) does not approach a finite limit as x → ∞, and there is no “right”
horizontal asymptote. What about “left” horizontal asymptotes? Well, we have
ex
x→−∞ 1 − e−x
=0
lim f (x) = lim
x→−∞
So y = 0 is the “left” horizontal asymptote of this function. What about
vertical asymptotes. As usual, we check to see where the function is undefined:
this happens whenever 1 − e−x is equal to zero. This happens for the value
x = 0. We should verify that this is a n asymptote of f (x):
e2x
x→0 1 − e−x
lim f (x) = lim
x→0
Notice that the numerator goes to 1; because 1 − e−x is positive for x > 0, it
follows that limx→0+ f (x) = ∞, and because 1 − e−x is negative for x < 0, we
have that limx→0− f (x) = −∞.
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