This document examines L'Hopital's rule for evaluating limits that result in indeterminate forms. It demonstrates how to apply the rule by taking derivatives of functions

1. 8.7 L’Hopital’s Rule revisited
Transcendental functions and applications

2. Recall that there are some functions that when finding limits
produce the indeterminate forms 0/0 or ∞/∞. When these come
up, they do not guarantee a limit exists, nor what that limit
is.
Some examples: 2x2 − 2 3x 2 − 2
lim lim 2
x →−1 x + 1 x →∞ 2 x + 1
Rewriting with algebraic techniques can handle some of them, for
example dividing out techniques, rationalizing denominators or
numerators, or using the squeeze theorem.

3. This rule is used incorrectly if you apply the quotient rule to
f(x)/g(x).

4. Let’s look at using L’Hôpital’s Rule on the examples we looked at
in the beginning.
2x2 − 2 Direct substitution achieves the indeterminate
lim form 0/0. Letting gand) = x + 1
(x
x →−1 x + 1
f ( x) = 2 x − 2
2
f ' ( x) 4x
= lim = lim = −4
x →−1 g ' ( x ) x →−1 1
Direct substitution achieves the indeterminate form
3x 2 − 2
lim 2 ∞/∞. Let f(x) be the numerator, and g(x) be
x →∞ 2 x + 1
the denominator.
3x 2 − 2 6x 6 3
lim 2 = lim = lim =
x →∞ 2 x + 1 x →∞ 4 x x →∞ 4 2

5. 0 , ±∞ , ∞ − ∞, 0 ⋅ ∞, 00 , 1∞ , and ∞ 0
The forms 0 ±∞
all have been identified as indeterminate. All but the first two would
have to be rewritten in quotient form before L’Hôpital’s Rule applies
There are similar forms that have been identified as
determinate. L’Hôpital’s Rule doesn’t apply.
∞+∞→∞ Limit is positive infinity
−∞ − ∞ → −∞ Limit is negative infinity
∞
0 →0 Limit is zero
−∞ Limit is positive infinity
0 →∞

6. Ex 1 p. 569 Indeterminant form 0/0
Evaluate e2 x − 1
lim x
x →0
Solution: Direct substitution yields 0/0
Taking the derivative of the numerator and dividing by the
derivative of the denominator, and finding limit of new
quotient:
2e 2 x
= lim
x →0 1
Now with direct substitution in this new but equivalent limit
according to L’Hopital’s Rule,
2e 2 x
lim 1 = 2
x →0

7. Ex 4 p570 Indeterminant form 0∙∞
Evaluate lim e− x x
x →∞
Because direct substitution yields 0∙∞, you need to do some kind of rewrite
to try to get in form yielding 0/0 or ∞/∞
x ∞
lim
x →∞ ex
=
∞
Apply L’Hopital’s Rule:
1
1
2 x = lim
lim
x →∞ ex x →∞ 2 xe x
As x increases, the denominator
gets large while the numerator
stays fixed. So the limit is 0.

8. 8.7c 574/ 6,12,15,18,21,24,30