MATH 116, LECTURE 22:
L’Hopital’s Rule
1
L’Hopital’s Rule
Another application of the derivative is in evaluating limits which have an
indeterminate form.
We saw several weeks ago that problems arise in evaluating rational
functions for which the limit, if we were to treat it as a simple number, gives
us one of the indeterminate forms
“0”
0
or
“∞”
.
∞
We encountered some different techniques for resolving these types of limits,
including factoring, rationalizing roots, dividing top and bottom by powers
of x, and the squeeze theorem. How about the following limit?
ln(x)
.
x→∞ x
lim
This is clearly an indeterminant form (∞/∞), but we cannot factor it, there
are no roots to rationalize, dividing by x doesn’t resolve anything, and there
is no clear candidate for an upper and lower bound on the term.
The following result, however, comes to our aid. It is an application of
the derivative, and is one of the most powerful tools for determining limits
of the form above.
Theorem 1.1 (L’Hopital’s Rule). Suppose that the functions f and g are
differentiable at x = a and that the limit as t → a of f (x)/g(x) has one of
the indeterminate forms “0/0” or “∞/∞”. Then we have
f (x)
f 0 (x)
= lim 0
.
x→a g(x)
x→a g (x)
lim
In other words, we can evaluate indeterminate limits of the forms “0/0”
and “∞/∞“ by taking the derivative of the top and bottom (separately!)
and evaluating that limit instead. It is not immediately obvious that this
1
should help us, but consider the example. We have
ln(x)
x→∞ x
lim
“
∞
”
∞
1
x
(L’Hopital’s)
1
1
= lim
= 0.
x→∞ x
There are a few notes worth making about the application of L’Hopital’s
Rule.
= lim
x→∞
1. The functions which we apply L’Hopital’s rule to are quotients, but
we do not apply the quotient rule to the derivates! The derivatives on
the top and bottom are evaluated separately.
2. Sometimes one application of L’Hopital’s rule will not be enough to
resolve the indeterminacy in the limit. We may have to perform it
multiple times (checking each time that the ratio has the proper form!).
3. L’Hopital’s rule only applies to indeterminate limits of the form “0/0”
and “∞/∞”. There are, however, many other indeterminate limit
forms. Many of them can be handled by simple manipulation:
(a) lim f (x)g(x) = “0 · ∞”
x→a
— Take the reciprocal of f (x) (or g(x)) to get a rational function
g(x)
∞
=“ ”
lim 1
∞
x→a
f (x)
or
0
f (x)
= “ ”.
lim 1
0
x→a
g(x)
Apply L’Hopital’s rule to this form.
(b) lim f (x)g(x) = “1∞ ”
x→a
— Take the limit of the natural logarithm to get
h
i
lim ln f (x)g(x) = lim [g(x) ln(f (x))] = “∞ · 0”.
x→a
x→a
Then apply the previous rule and remember to take the exponent
of the answer!
(c) lim f (x)g(x) = “00 ”
x→a
— Again, take the limit of the natural logarithm to get
h
i
lim ln f (x)g(x) = lim [g(x) ln(f (x))] = −“0 · ∞”
x→a
x→a
then apply the first rule.
2
Example 1: Evaluate
lim
t→0
sin(t)
t
using L’Hopital’s rule.
Solution: We already know that this limit evaluates to one by an application of the squeeze theorem. Let’s see if we can get the same result using
L’Hopital’s rule. We have
sin(t)
t→0
t
cos(t)
= lim
t→0
1
= 1.
lim
0
“ ”
0
Example 2: Evaluate
2x2 − 5
x→∞ 3x2 + 7x + 15
lim
using L’Hopital’s rule.
Solution: We have
2x2 − 5
x→∞ 3x2 + 7x + 15
4x
= lim
x→0 6x + 7
4
= lim
x→0 6
2
= .
3
lim
Example 3: Evaluate
sin(x) − x
.
x→0
x3
lim
3
∞
”
∞
∞
“ ”
∞
“
Solution: We have
sin(x) − x
x→0
x3
cos(x) − 1
= lim
x→0
3x2
− sin(x)
= lim
x→0
6x
− cos(x)
= lim
x→0
6
1
=− .
6
Example 4: Evaluate
1 x
lim 1 −
.
x→∞
x
lim
0
“ ”
0
0
“ ”
0
0
“ ”
0
Solution: We have
1 x
lim 1 −
= “1∞ ”.
x→∞
x
We need to rearrange this to get it into a form “0/0” or “∞/∞”. To resolve
the exponent, we take the natural logarithm of both sides to get
1 x
1
lim ln 1 −
.
= lim x ln 1 −
x→∞
x→∞
x
x
This is a limit of the form ∞ · 0, so we divide by the reciprocal of one of the
terms. We have
ln 1 − x1
0
lim
“ ”
1
x→∞
0
x
1
1/ 1 − x1
2
x
= lim
x→∞
− x12
1
= − lim
x→∞ 1 − 1
x
= −1.
We recall this is the limit of natural logarithm of the limit we want, so
we really have
1 x
1
lim 1 −
= e−1 = .
x→∞
x
e
4