Lesson17 -indeterminate_forms_and_l_hopitals_rule_021_handout

Section 3.7
Indeterminate Forms and L’Hôpital’s
Rule
V63.0121.021, Calculus I
New York University
November 4, 2010
Announcements
Quiz 3 in recitation this week on Sections 2.6, 2.8, 3.1, and 3.2
Announcements
Quiz 3 in recitation this
week on Sections 2.6, 2.8,
3.1, and 3.2
V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 2 / 34
Objectives
Know when a limit is of
indeterminate form:
indeterminate quotients:
0/0, ∞/∞
indeterminate products:
0 × ∞
indeterminate differences:
∞ − ∞
indeterminate powers: 00
,
∞0
, and 1∞
Resolve limits in
indeterminate form
Notes
Notes
Notes
1
Section 3.7 : L’Hôpital’s RuleV63.0121.021, Calculus I November 4, 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
Relative Rates of Growth
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
Limits of Rational Functions revisited
Example
Find lim
x→∞
5x2 + 3x − 1
3x2 + 7x + 27
if it exists.
Solution
Using L’Hˆopital:
lim
x→∞
5x2 + 3x − 1
3x2 + 7x + 27
H
= lim
x→∞
10x + 3
6x + 7
H
= lim
x→∞
10
6
=
5
3
Notes
Notes
Notes
6

Limits of Rational Functions revisited II
Example
Find lim
x→∞
5x2 + 3x − 1
7x + 27
if it exists.
Solution
lim
x→∞
5x2 + 3x − 1
7x + 27
H
= lim
x→∞
10x + 3
7
= ∞
Limits of Rational Functions revisited III
Example
Find lim
x→∞
4x + 7
3x2 + 7x + 27
if it exists.
Solution
lim
x→∞
4x + 7
3x2 + 7x + 27
H
= lim
x→∞
4
6x + 7
= 0
Limits of Rational Functions
Fact
Let f (x) and g(x) be polynomials of degree p and q.
If p > q, then lim
x→∞
f (x)
g(x)
= ∞
If p < q, then lim
x→∞
f (x)
g(x)
= 0
If p = q, then lim
x→∞
f (x)
g(x)
is the ratio of the leading coeﬃcients of f
and g.
Notes
Notes
Notes
7

Exponential versus geometric growth
Example
Find lim
x→∞
ex
x2
, if it exists.
Solution
We have
lim
x→∞
ex
x2
H
= lim
x→∞
ex
2x
H
= lim
x→∞
ex
2
= ∞.
Example
What about lim
x→∞
ex
x3
?
Answer
Exponential versus fractional powers
Example
Find lim
x→∞
ex
√
x
, if it exists.
Solution (without L’Hôpital)
We have for all x > 1, x1/2
< x1
, so
ex
x1/2
>
ex
x
The right hand side tends to ∞, so the left-hand side must, too.
Solution (with L’Hôpital)
lim
x→∞
ex
√
x
= lim
x→∞
ex
1
2x−1/2
= lim
x→∞
2
√
xex
= ∞
Exponential versus any power
Theorem
Let r be any positive number. Then
lim
x→∞
ex
xr
= ∞.
Proof.
If r is a positive integer, then apply L’Hôpital’s rule r times to the
fraction. You get
lim
x→∞
ex
xr
H
= . . .
H
= lim
x→∞
ex
r!
= ∞.
If r is not an integer, let m be the smallest integer greater than r. Then if
x > 1, xr
< xm
, so
ex
xr
>
ex
xm
. The right-hand side tends to ∞ by the
previous step.
Notes
Notes
Notes
8

Any exponential versus any power
Theorem
Let a > 1 and r > 0. Then
lim
x→∞
ax
xr
= ∞.
Proof.
If r is a positive integer, we have
lim
x→∞
ax
xr
H
= . . .
H
= lim
x→∞
(ln a)r ax
r!
= ∞.
If r isn’t an integer, we can compare it as before.
So even lim
x→∞
(1.00000001)x
x100000000
= ∞!
Logarithmic versus power growth
Theorem
Let r be any positive number. Then
lim
x→∞
ln x
xr
= 0.
Proof.
One application of L’Hˆopital’s Rule here suﬃces:
lim
x→∞
ln x
xr
H
= lim
x→∞
1/x
rxr−1
= lim
x→∞
1
rxr
= 0.
Outline
Notes
Notes
Notes
9

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
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’Hôpital:
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
.
Notes
Notes
Notes
10

Another indeterminate power limit
Example
lim
x→0
(3x)4x
Solution
Summary
Form Method
0
0 L’Hôpital’s rule directly
∞
∞ L’Hôpital’s rule directly
0 · ∞ jiggle to make 0
0 or ∞
∞ .
∞ − ∞ combine to make an indeterminate product or quotient
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
11

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