Lesson 17: Indeterminate Forms and L'Hopital's Rule (Section 041 handout)

Section 3.7
Indeterminate Forms and L’Hôpital’s
Rule
V63.0121.041, Calculus I
New York University
November 3, 2010
Announcements
Announcements
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 2 / 24
Objectives
Know when a limit is of
indeterminate form:
indeterminate quotients:
0/0, ∞/∞
indeterminate products:
0 × ∞
indeterminate differences:
∞ − ∞
indeterminate powers: 00
,
∞0
, and 1∞
Notes
Notes
Notes
1
Section 3.7 : L’Hôpital’s RuleV63.0121.041, Calculus I November 3, 2010

Experiments with funny limits
lim
x→0
sin2
x
x
= 0
lim
x→0
x
sin2
x
does not exist
lim
x→0
sin2
x
sin(x2)
= 1
lim
x→0
sin 3x
sin x
= 3
All of these are of the form
0
0
, and since we can get different answers in
different cases, we say this form is indeterminate.
Recall
Recall the limit laws from Chapter 2.
Limit of a sum is the sum of the limits
Limit of a difference is the difference of the limits
Limit of a product is the product of the limits
Limit of a quotient is the quotient of the limits ... whoops! This is
true as long as you don’t try to divide by zero.
More about dividing limits
We know dividing by zero is bad.
Most of the time, if an expression’s numerator approaches a finite
number and denominator approaches zero, the quotient approaches
some kind of infinity. For example:
lim
x→0+
1
x
= +∞ lim
x→0−
cos x
x3
= −∞
An exception would be something like
lim
x→∞
1
1
x sin x
= lim
x→∞
x csc x.
which does not exist and is not infinite.
Even less predictable: numerator and denominator both go to zero.
Notes
Notes
Notes
2

Language Note
It depends on what the meaning of the word “is” is
Be careful with the language
here. We are not saying that
the limit in each case “is”
0
0
, and therefore nonexistent
because this expression is
undefined.
The limit is of the form
0
0
,
which means we cannot
evaluate it with our limit
laws.
Indeterminate forms are like Tug Of War
Which side wins depends on which side is stronger.
Outline
L’Hôpital’s Rule
Other Indeterminate Limits
Indeterminate Products
Indeterminate Differences
Indeterminate Powers
Notes
Notes
Notes
3

The Linear Case
Question
If f and g are lines and f (a) = g(a) = 0, what is
lim
x→a
f (x)
g(x)
?
Solution
The functions f and g can be written in the form
f (x) = m1(x − a)
g(x) = m2(x − a)
So
f (x)
g(x)
=
m1
m2
The Linear Case, Illustrated
x
y
y = f (x)
y = g(x)
a
x
f (x)
g(x)
f (x)
g(x)
=
f (x) − f (a)
g(x) − g(a)
=
(f (x) − f (a))/(x − a)
(g(x) − g(a))/(x − a)
=
m1
m2
What then?
But what if the functions aren’t linear?
Can we approximate a function near a point with a linear function?
What would be the slope of that linear function? The derivative!
Notes
Notes
Notes
4

Theorem of the Day
Theorem (L’Hopital’s Rule)
Suppose f and g are differentiable functions and g (x) = 0 near a (except
possibly at a). Suppose that
lim
x→a
f (x) = 0 and lim
x→a
g(x) = 0
or
lim
x→a
f (x) = ±∞ and lim
x→a
g(x) = ±∞
Then
lim
x→a
f (x)
g(x)
= lim
x→a
f (x)
g (x)
,
if the limit on the right-hand side is finite, ∞, or −∞.
Meet the Mathematician: L’Hôpital
wanted to be a military
man, but poor eyesight
forced him into math
did some math on his own
(solved the “brachistocrone
problem”)
paid a stipend to Johann
Bernoulli, who proved this
theorem and named it after
him! Guillaume Fran¸cois Antoine,
Marquis de L’Hôpital
(French, 1661–1704)
Revisiting the previous examples
Example
lim
x→0
sin2
x
x
H
= lim
x→0
2 sin x
sin x → 0
cos x
1
= 0
Example
lim
x→0
sin2
x
numerator → 0
sin x2
denominator → 0
H
= lim
x→0
¡2 sin x cos x
numerator → 0
(cos x2) (¡2x
denominator → 0
)
H
= lim
x→0
cos2 x − sin2
x
numerator → 1
cos x2 − 2x2 sin(x2)
denominator → 1
= 1
Example
lim
x→0
sin 3x
sin x
H
= lim
x→0
3 cos 3x
cos x
= 3.
Notes
Notes
Notes
5

Beware of Red Herrings
Example
Find
lim
x→0
x
cos x
Solution
Outline
L’Hôpital’s Rule
Other Indeterminate Limits
Indeterminate Products
Indeterminate Differences
Indeterminate Powers
Indeterminate products
Example
Find
lim
x→0+
√
x ln x
This limit is of the form 0 · (−∞).
Solution
Jury-rig the expression to make an indeterminate quotient. Then apply
L’Hôpital’s Rule:
lim
x→0+
√
x ln x = lim
x→0+
ln x
1/
√
x
H
= lim
x→0+
x−1
−1
2x−3/2
= lim
x→0+
−2
√
x = 0
Notes
Notes
Notes
6

Indeterminate diﬀerences
Example
lim
x→0+
1
x
− cot 2x
This limit is of the form ∞ − ∞.
Solution
Indeterminate powers
Example
Find lim
x→0+
(1 − 2x)1/x
Take the logarithm:
ln lim
x→0+
(1 − 2x)1/x
= lim
x→0+
ln (1 − 2x)1/x
= lim
x→0+
ln(1 − 2x)
x
This limit is of the form
0
0
, so we can use L’Hˆopital:
lim
x→0+
ln(1 − 2x)
x
H
= lim
x→0+
−2
1−2x
1
= −2
This is not the answer, it’s the log of the answer! So the answer we want
is e−2
.
Another indeterminate power limit
Example
lim
x→0
(3x)4x
Solution
Notes
Notes
Notes
7

Summary
Form Method
0
0 L’Hôpital’s rule directly
∞
∞ L’Hôpital’s rule directly
0 · ∞ jiggle to make 0
0 or ∞
∞ .
∞ − ∞ factor to make an indeterminate product
00 take ln to make an indeterminate product
∞0 ditto
1∞ ditto
Final Thoughts
L’Hôpital’s Rule only works on indeterminate quotients
Luckily, most indeterminate limits can be transformed into
indeterminate quotients
L’Hôpital’s Rule gives wrong answers for non-indeterminate limits!
Notes
Notes
Notes
8

Lesson 17: Indeterminate Forms and L'Hopital's Rule (Section 041 handout)

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