10/27/10 L’HOPITAL’S RULE
ALEX KRUCKMAN
(1) L’Hopital’s Rule is a general method for dealing with limits of quotients of functions which would
∞
evaluate to 00 by direct substitution. 00 , along with ∞
∞ , −∞ , etc. are called indeterminate forms.
0
Note that ∞
and ∞
0 are NOT indeterminate forms. Why?
(2) L’Hopital’s Rule gives quick answers for limits which used to be tricky. For example, we showed
that
sin(x)
lim
=1
x→0
x
by a tricky geometric argument. With L’Hopital’s rule,
sin(x)
cos(x)
= lim
= 1.
x→0
x
1
(3) We can also apply L’Hopital’s Rule for limits at infinity. For example,
cos x1 − x12
sin x1
= lim lim
,
x→∞
x→∞ tan−1 1
1
1
x
− x2
1+ 1
lim
x→0
x2
since limx→∞ sin( x1 ) = limx→∞ tan−1 ( x1 ) = 0. This is
1
1
lim cos
1 + 2 = 1.
x→∞
x
x
(4) L’Hopital’s rule can help answer questions about continuity and differentiability. The function
ex −1
is not defined at 0. But can we extend it to a function which is continuous everywhere? If
x
its graph looks like a continuous curve with a hole, we can do this by filling the hole. Check the
limit as x goes to 0:
ex − 1
ex
lim
= lim
= 1.
x→0
x→0 1
x
So the function
( x
e −1
x 6= 0
x
f (x) =
1
x=0
is continuous. Is it differentiable at 0? Let’s try to compute its derivative.
x
e −1
−1
ex − x − 1
f (x) − f (0)
= lim x
= lim
.
x→0
x→0
x→0
x−0
x
x2
Once again, we have an indeterminate form, so we can apply L’Hopital’s Rule. We get
f 0 (0) = lim
ex − 1
ex
1
= lim
= .
x→0 2x
x→0 2
2
Note that we had to apply L’Hopital’s Rule twice in this example.
(5) L’Hopital’s Rule also applies to other indeterminate forms which are not 00 . For example,
lim
lim
x→0+
ln(x)
1
x
= lim
x→0+
since limx→=0+ = −∞, and limx→0+
1
x
− x12
1
x
= lim −
x→0+
= ∞.
1
x2
= lim −x = 0,
x
x→0+
2
ALEX KRUCKMAN
(6) Sometimes a limit will be indeterminate, but will not appear in the nice form 00 or ∞
∞ . In these
cases L’Hopital’s Rule may still apply, after some figiting to get the limit into a nice form.
For example, take limx→0+ (x + 1)cot(x) . As x approaches 0, the base approaches 1, but the
exponent approaches ∞. We can’t determine what happens without more information. Since
there’s a tricky exponential, we think to take logarithms.
y = (x + 1)cot(x)
ln(y) = cot(x) ln(x + 1)
cos(x) ln(x + 1)
=
sin(x)
This looks like logarithmic differentiation, but we’re looking for a limit here, not a derivative.
But notice that the right side is now a quotient. Starting to look like L’Hopital’s Rule? Let’s try
taking the limit.
lim ln(y) =
x→0
lim
x→0
cos(x) ln(x + 1)
sin(x)
1
− sin(x) ln(x + 1) + cos(x) x+1
x→0
cos(x)
1
=
= 1.
1
Magic! But we’ve computed the limit of ln(y), and we wanted the limit of y. Solution?
Exponentiate. limx→0 y = limx→0 eln(y) = elimx→0 ln(y) = e1 = e. Weird how e pops up everywhere,
right?
=
lim