L’Hopital’s rule
∞
0
There are seven so-called “indeterminate forms” in math: 00 , 00 , ∞
∞ , ∞ − ∞, 1 , ∞ , 0 · ∞. L’Hopital’s
rule helps evaluate limits when you get an indeterminate form. For example, consider
sin(x)
.
x
lim
x→0
As x → 0, both sin(x) and x approach 0, so this limit becomes 0/0. This is certainly not helpful. So we use
L’Hopital’s Rule:
L’Hopital: Suppose limx→c
f (x)
g(x)
=
0
0
or
∞
∞,
where c is allowed to be ±∞. Then
f (x)
f 0 (x)
= lim 0
,
x→c g(x)
x→c g (x)
lim
provided the right hand side exists.
So L’Hopital’s rule is simple: provided we have one of those two forms, we just take the derivative of
the top and the bottom (separately; note no quotient rule is needed there) and then re-evaulate the limit.
So for example, since the limit above gives 0/0, we see
(sin(x))0
cos(x)
sin(x)
= lim
= lim
= 1.
x→0
x→0
x→0
x
(x)0
1
lim
Exercise: Use L’Hopital’s rule to show:
x2 + 4
=0
x→∞
ex
lim
ex − 1
= 1.
x→0
x
and
lim
Suppose your limit became 0 · ∞ instead? In that case, you need to somehow manipulate the limit to
get it into either 0/0 or ∞/∞ form.
1. CASE 1: 0 · ∞. Suppose we had limx→−∞ xex . This is of the form 0 · ∞. To get it into the desired
1
form, we write ex = e−x
, so that
x
lim xex = lim
x→−∞
x→−∞ e−x
1
= 0,
x→−∞ −e−x
= lim
where we could use L’Hopital’s rule at the end since we had an ∞/∞ form. Similarly, if we wanted to
evaluate
lim x ln(x),
x→0
then we could write x =
1
1
x
, so that
lim x ln(x) = lim
x→0
x→0
ln(x)
1
x
1
x
x→0 − 12
x
= lim
= lim −x = 0.
x→0
So when you have an integral of this type, transform the integral as above to get either 0/0 or ∞/∞.
2. CASE 2: 00 , 1∞ , ∞0 . We deal with these cases similarly. Consider
lim xx .
x→0
1
Suppose we write y = xx . Then ln(y) = x ln(x) (log rules). Therefore
lim ln(y) = lim x ln(x) = 0
x→0
x→0
(as in Case 1). But since eln(y) = y = xx , we find
lim xx = lim y = lim eln(y) = elimx→0 y = e0 = 1.
x→0
x→0
x→0
So the trick is to take logs, evaluate the limit, and then exponentiate your answer to get the solution
to your original limit. Let’s do another:
x
1
.
lim 1 +
x→∞
x
x
So if y = 1 + x1 , ln(y) = x ln 1 + x1 . Now
1
lim x ln 1 +
x→∞
x
is a Case 1 integral. Practice that case by doing this limit, and see that you get 1. Therefore the
solution to your original limit is e1 = e.
2