Given that we’ve been computing asymptotes, we often run into a situation
where the limit of the ratio of two functions is undefined. As a simple example,
consider the function
5x
f (x) =
.
x−3
We want to consider the limit as x approaches 3. Notice that, if we were just
to plug in 3, the numerator is equal to 15 and the denominator is equal to 0. It
turns out that if the numerator is equal to zero and the denominator isn’t, then
f (x) will have a vertical asymptote. So we immediately know that f (x) has a
vertical asymptote at x = 3.
On the other hand, if both the numerator and denominator are equal to
zero, it’s less clear what happens. We can have a hole or a vertical asymptote
(or maybe something even worse). In the example presented last week, we
were able to factor the numerator and denominator in order to cancel out the
common root. Unfortunately, this isn’t always possible. For example, consider
the function
ex − 1
.
f (x) =
x2
Let’s take the limit as x approaches zero:
ex − 1
.
x→0
x
lim f (x) = lim
x→0
Notice that if we just plug in zero, we get that the numerator and denominator
both go to zero. However, as we can see from the geogebra file, the limit is equal
to 1.
We would like a method to algebraically find the limit in cases such as this.
The idea here is that we approximate the numerator and denominator by their
tangent lines: Clearly the tangent line to y = x is y = x. The tangent line to
ex − 1 at x = 0 can be found by taking the derivative: the slope is seen to be
1, and e0 − 1 = 0, so the tangent line to the numerator is y = x. Therefore, the
limit should be the same as
x
lim
x→0 x
which is equal to 1.
Notice that the above tangent lines will always be of the form y = m(x − c),
where c is the point where the limit is being taken, for some m because we’re
assuming the numerator and denominator are both equal to zero at x = c.
Therefore, the above limit will always be the same as the ratio of the derivatives.
This suggests the following rule:
(x)
Theorem 0.1 (L’Hôpital’s rule). Suppose that h(x) := fg(x)
is a quotient of
two functions. Suppose that limx→c f (x) and limx→c g(x) are both equal to zero.
Then we have that
f ′ (x)
.
lim h(x) = lim ′
x→c
x→c g (x)
This rule works even if c = ∞ or c = −∞.
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Notice that this version of L’Hôpital’s rule requires that both the numerator
and denominator go to zero. There is another version of L’Hôpital’s rule that
is useful for finding horizontal asymptotes:
(x)
is a
Theorem 0.2 (L’Hôpital’s rule, version 2). Suppose that h(x) := fg(x)
quotient of two functions. Suppose that limx→c f (x) and limx→c g(x) are equal
to ±∞ (you don’t need both of them to be ∞ or both of them to be −∞). Then
we have that
f ′ (x)
lim h(x) = lim ′
x→c
x→c g (x)
This rule works even if c = ∞ or c = −∞.
Example 0.3. Find the limit as x → 1 of
ln(x2 − 2x + 2)
.
x−1
Directly plugging in x = 1 gives
ln(1)
0 ,
which is 00 , so L’Hôpital’s rule applies:
ln(x2 − 2x + 2)
= lim
x→1
x→1
x−1
lim
1
x2 −2x+2 (2x
− 2)
1
2x − 2
= lim 2
x→1 x − 2x + 2
=0
so the limit is zero. The numerator is going to zero much faster than x − 1 does.
What happens if we make the denominator smaller?
Example 0.4. Find the limit as x → 1 of
ln(x2 − 2x + 2)
.
(x − 1)2
This time, when we take the derivative, we get:
ln(x2 − 2x + 2)
= lim
x→1
x→1
(x − 1)2
lim
1
x2 −2x+2 (2x
− 2)
(2x − 2)
2x − 2
= lim
x→1 (2x − 2)(x2 − 2x + 2)
At this stage, it should immediately be clear that we can cancel out the 2x− 2’s,
but the point I want to make is that we can also just use L’Hôpital’s rule again:
2x − 2
ln(x2 − 2x + 2)
= lim
x→1 (2x − 2)(x2 − 2x + 2)
x→1
(x − 1)2
2
= lim
x→1 2(x2 − 2x + 2) + (2x − 2)(2x − 2)
2
=
2(1) + 0
=1
lim
2
Therefore, the limit is 1.
Let’s consider another example
Example 0.5. Find the limit as x → 0 of
f (x) = x ln x.
This time, when we plug in zero, we seem to get 0 × (−∞), which doesn’t
mean anything. So we rewrite the expression as
f (x) =
ln x
1/x
Now, the numerator is going to −∞, and the denominator is going to ∞ (here,
x > 0 by the domain restriction on f ). So we can apply L’Hôpital’s rule:
ln x
1/x
= lim
x→0 1/x
x→0 −1/x2
−x2
= lim
x→0 x
= lim −x
lim
x→0
=0
So the limit is zero.
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