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Section 3.7Indeterminate Forms and L’Hôpital’s
Rule
V63.0121.041, Calculus I
New York University
November 3, 2010
Announcements
I Quiz 3 in recitation this week on Sections 2.6, 2.8, 3.1, and 3.2
. . . . . .
. . . . . .
Announcements
I Quiz 3 in recitation thisweek on Sections 2.6, 2.8,3.1, and 3.2
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 2 / 34
. . . . . .
Objectives
I Know when a limit is ofindeterminate form:
I indeterminate quotients:0/0, ∞/∞
I indeterminate products:0×∞
I indeterminatedifferences: ∞−∞
I indeterminate powers:00, ∞0, and 1∞
I Resolve limits inindeterminate form
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 3 / 34
. . . . . .
Experiments with funny limits
I limx→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different answers
in different cases, we say this form is indeterminate.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34
. . . . . .
Experiments with funny limits
I limx→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different answers
in different cases, we say this form is indeterminate.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34
. . . . . .
Experiments with funny limits
I limx→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different answers
in different cases, we say this form is indeterminate.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34
. . . . . .
Experiments with funny limits
I limx→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different answers
in different cases, we say this form is indeterminate.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34
. . . . . .
Experiments with funny limits
I limx→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different answers
in different cases, we say this form is indeterminate.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34
. . . . . .
Experiments with funny limits
I limx→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different answers
in different cases, we say this form is indeterminate.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34
. . . . . .
Experiments with funny limits
I limx→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different answers
in different cases, we say this form is indeterminate.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34
. . . . . .
Experiments with funny limits
I limx→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different answers
in different cases, we say this form is indeterminate.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34
. . . . . .
Experiments with funny limits
I limx→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different answers
in different cases, we say this form is indeterminate.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 4 / 34
. . . . . .
Recall
Recall the limit laws from Chapter 2.
I Limit of a sum is the sum of the limits
I Limit of a difference is the difference of the limitsI Limit of a product is the product of the limitsI 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.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 5 / 34
. . . . . .
Recall
Recall the limit laws from Chapter 2.
I Limit of a sum is the sum of the limitsI Limit of a difference is the difference of the limits
I Limit of a product is the product of the limitsI 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.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 5 / 34
. . . . . .
Recall
Recall the limit laws from Chapter 2.
I Limit of a sum is the sum of the limitsI Limit of a difference is the difference of the limitsI Limit of a product is the product of the limits
I Limit of a quotient is the quotient of the limits ... whoops! This istrue as long as you don’t try to divide by zero.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 5 / 34
. . . . . .
Recall
Recall the limit laws from Chapter 2.
I Limit of a sum is the sum of the limitsI Limit of a difference is the difference of the limitsI Limit of a product is the product of the limitsI 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.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 5 / 34
. . . . . .
More about dividing limits
I We know dividing by zero is bad.I Most of the time, if an expression’s numerator approaches a finite
number and denominator approaches zero, the quotientapproaches some kind of infinity. For example:
limx→0+
1x= +∞ lim
x→0−cos xx3
= −∞
I An exception would be something like
limx→∞
11x sin x
= limx→∞
x csc x.
which does not exist and is not infinite.I Even less predictable: numerator and denominator both go to
zero.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 6 / 34
. . . . . .
More about dividing limits
I We know dividing by zero is bad.I Most of the time, if an expression’s numerator approaches a finite
number and denominator approaches zero, the quotientapproaches some kind of infinity. For example:
limx→0+
1x= +∞ lim
x→0−cos xx3
= −∞
I An exception would be something like
limx→∞
11x sin x
= limx→∞
x csc x.
which does not exist and is not infinite.
I Even less predictable: numerator and denominator both go tozero.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 6 / 34
. . . . . .
More about dividing limits
I We know dividing by zero is bad.I Most of the time, if an expression’s numerator approaches a finite
number and denominator approaches zero, the quotientapproaches some kind of infinity. For example:
limx→0+
1x= +∞ lim
x→0−cos xx3
= −∞
I An exception would be something like
limx→∞
11x sin x
= limx→∞
x csc x.
which does not exist and is not infinite.I Even less predictable: numerator and denominator both go to
zero.V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 6 / 34
. . . . . .
Language NoteIt depends on what the meaning of the word “is" is
I Be careful with thelanguage here. We are notsaying that the limit in each
case “is”00, and therefore
nonexistent because thisexpression is undefined.
I The limit is of the form00,
which means we cannotevaluate it with our limitlaws.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 7 / 34
. . . . . .
Indeterminate forms are like Tug Of War
Which side wins depends on which side is stronger.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 8 / 34
. . . . . .
Outline
L’Hôpital’s Rule
Relative Rates of Growth
Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 9 / 34
. . . . . .
The Linear Case
QuestionIf f and g are lines and f(a) = g(a) = 0, what is
limx→a
f(x)g(x)
?
SolutionThe functions f and g can be written in the form
f(x) = m1(x− a)g(x) = m2(x− a)
Sof(x)g(x)
=m1m2
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 10 / 34
. . . . . .
The Linear Case
QuestionIf f and g are lines and f(a) = g(a) = 0, what is
limx→a
f(x)g(x)
?
SolutionThe functions f and g can be written in the form
f(x) = m1(x− a)g(x) = m2(x− a)
Sof(x)g(x)
=m1m2
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 10 / 34
. . . . . .
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)
=m1m2
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 11 / 34
. . . . . .
What then?
I But what if the functions aren’t linear?
I Can we approximate a function near a point with a linear function?I What would be the slope of that linear function? The derivative!
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 12 / 34
. . . . . .
What then?
I But what if the functions aren’t linear?I Can we approximate a function near a point with a linear function?
I What would be the slope of that linear function? The derivative!
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 12 / 34
. . . . . .
What then?
I But what if the functions aren’t linear?I Can we approximate a function near a point with a linear function?I What would be the slope of that linear function?
The derivative!
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 12 / 34
. . . . . .
What then?
I But what if the functions aren’t linear?I Can we approximate a function near a point with a linear function?I What would be the slope of that linear function? The derivative!
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 12 / 34
. . . . . .
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
limx→a
f(x) = 0 and limx→a
g(x) = 0
or
limx→a
f(x) = ±∞ and limx→a
g(x) = ±∞
Thenlimx→a
f(x)g(x)
= limx→a
f′(x)g′(x)
,
if the limit on the right-hand side is finite, ∞, or −∞.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 13 / 34
. . . . . .
Meet the Mathematician: L'H_pital
I wanted to be a militaryman, but poor eyesightforced him into math
I did some math on his own(solved the “brachistocroneproblem”)
I paid a stipend to JohannBernoulli, who proved thistheorem and named it afterhim! Guillaume François Antoine,
Marquis de L’Hôpital(French, 1661–1704)
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 14 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→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
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 15 / 34
. . . . . .
Another Example
Example
Findlimx→0
xcos x
SolutionThe limit of the denominator is 1, not 0, so L’Hôpital’s rule does notapply. The limit is 0.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 16 / 34
. . . . . .
Beware of Red Herrings
Example
Findlimx→0
xcos x
SolutionThe limit of the denominator is 1, not 0, so L’Hôpital’s rule does notapply. The limit is 0.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 16 / 34
. . . . . .
Outline
L’Hôpital’s Rule
Relative Rates of Growth
Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 17 / 34
. . . . . .
Limits of Rational Functions revisited
Example
Find limx→∞
5x2 + 3x− 13x2 + 7x+ 27
if it exists.
SolutionUsing L’Hôpital:
limx→∞
5x2 + 3x− 13x2 + 7x+ 27
H= lim
x→∞
10x+ 36x+ 7
H= lim
x→∞
106
=53
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34
. . . . . .
Limits of Rational Functions revisited
Example
Find limx→∞
5x2 + 3x− 13x2 + 7x+ 27
if it exists.
SolutionUsing L’Hôpital:
limx→∞
5x2 + 3x− 13x2 + 7x+ 27
H= lim
x→∞
10x+ 36x+ 7
H= lim
x→∞
106
=53
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34
. . . . . .
Limits of Rational Functions revisited
Example
Find limx→∞
5x2 + 3x− 13x2 + 7x+ 27
if it exists.
SolutionUsing L’Hôpital:
limx→∞
5x2 + 3x− 13x2 + 7x+ 27
H= lim
x→∞
10x+ 36x+ 7
H= lim
x→∞
106
=53
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34
. . . . . .
Limits of Rational Functions revisited
Example
Find limx→∞
5x2 + 3x− 13x2 + 7x+ 27
if it exists.
SolutionUsing L’Hôpital:
limx→∞
5x2 + 3x− 13x2 + 7x+ 27
H= lim
x→∞
10x+ 36x+ 7
H= lim
x→∞
106
=53
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34
. . . . . .
Limits of Rational Functions revisited
Example
Find limx→∞
5x2 + 3x− 13x2 + 7x+ 27
if it exists.
SolutionUsing L’Hôpital:
limx→∞
5x2 + 3x− 13x2 + 7x+ 27
H= lim
x→∞
10x+ 36x+ 7
H= lim
x→∞
106
=53
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 18 / 34
. . . . . .
Limits of Rational Functions revisited II
Example
Find limx→∞
5x2 + 3x− 17x+ 27
if it exists.
SolutionUsing L’Hôpital:
limx→∞
5x2 + 3x− 17x+ 27
H= lim
x→∞
10x+ 37
= ∞
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 19 / 34
. . . . . .
Limits of Rational Functions revisited II
Example
Find limx→∞
5x2 + 3x− 17x+ 27
if it exists.
SolutionUsing L’Hôpital:
limx→∞
5x2 + 3x− 17x+ 27
H= lim
x→∞
10x+ 37
= ∞
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 19 / 34
. . . . . .
Limits of Rational Functions revisited II
Example
Find limx→∞
5x2 + 3x− 17x+ 27
if it exists.
SolutionUsing L’Hôpital:
limx→∞
5x2 + 3x− 17x+ 27
H= lim
x→∞
10x+ 37
= ∞
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 19 / 34
. . . . . .
Limits of Rational Functions revisited II
Example
Find limx→∞
5x2 + 3x− 17x+ 27
if it exists.
SolutionUsing L’Hôpital:
limx→∞
5x2 + 3x− 17x+ 27
H= lim
x→∞
10x+ 37
= ∞
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 19 / 34
. . . . . .
Limits of Rational Functions revisited III
Example
Find limx→∞
4x+ 73x2 + 7x+ 27
if it exists.
SolutionUsing L’Hôpital:
limx→∞
4x+ 73x2 + 7x+ 27
H= lim
x→∞
46x+ 7
= 0
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 20 / 34
. . . . . .
Limits of Rational Functions revisited III
Example
Find limx→∞
4x+ 73x2 + 7x+ 27
if it exists.
SolutionUsing L’Hôpital:
limx→∞
4x+ 73x2 + 7x+ 27
H= lim
x→∞
46x+ 7
= 0
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 20 / 34
. . . . . .
Limits of Rational Functions revisited III
Example
Find limx→∞
4x+ 73x2 + 7x+ 27
if it exists.
SolutionUsing L’Hôpital:
limx→∞
4x+ 73x2 + 7x+ 27
H= lim
x→∞
46x+ 7
= 0
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 20 / 34
. . . . . .
Limits of Rational Functions revisited III
Example
Find limx→∞
4x+ 73x2 + 7x+ 27
if it exists.
SolutionUsing L’Hôpital:
limx→∞
4x+ 73x2 + 7x+ 27
H= lim
x→∞
46x+ 7
= 0
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 20 / 34
. . . . . .
Limits of Rational Functions
FactLet f(x) and g(x) be polynomials of degree p and q.
I If p > q, then limx→∞
f(x)g(x)
= ∞
I If p < q, then limx→∞
f(x)g(x)
= 0
I If p = q, then limx→∞
f(x)g(x)
is the ratio of the leading coefficients of f
and g.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 21 / 34
. . . . . .
Exponential versus geometric growth
Example
Find limx→∞
ex
x2, if it exists.
SolutionWe have
limx→∞
ex
x2H= lim
x→∞
ex
2xH= lim
x→∞
ex
2= ∞.
Example
What about limx→∞
ex
x3?
AnswerStill ∞. (Why?)
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34
. . . . . .
Exponential versus geometric growth
Example
Find limx→∞
ex
x2, if it exists.
SolutionWe have
limx→∞
ex
x2H= lim
x→∞
ex
2xH= lim
x→∞
ex
2= ∞.
Example
What about limx→∞
ex
x3?
AnswerStill ∞. (Why?)
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34
. . . . . .
Exponential versus geometric growth
Example
Find limx→∞
ex
x2, if it exists.
SolutionWe have
limx→∞
ex
x2H= lim
x→∞
ex
2xH= lim
x→∞
ex
2= ∞.
Example
What about limx→∞
ex
x3?
AnswerStill ∞. (Why?)
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34
. . . . . .
Exponential versus geometric growth
Example
Find limx→∞
ex
x2, if it exists.
SolutionWe have
limx→∞
ex
x2H= lim
x→∞
ex
2xH= lim
x→∞
ex
2= ∞.
Example
What about limx→∞
ex
x3?
AnswerStill ∞. (Why?)
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 22 / 34
. . . . . .
Exponential versus fractional powers
Example
Find limx→∞
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)
limx→∞
ex√x= lim
x→∞
ex
12x−1/2
= limx→∞
2√xex = ∞
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 23 / 34
. . . . . .
Exponential versus fractional powers
Example
Find limx→∞
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)
limx→∞
ex√x= lim
x→∞
ex
12x−1/2
= limx→∞
2√xex = ∞
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 23 / 34
. . . . . .
Exponential versus fractional powers
Example
Find limx→∞
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)
limx→∞
ex√x= lim
x→∞
ex
12x−1/2
= limx→∞
2√xex = ∞
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 23 / 34
. . . . . .
Exponential versus any power
TheoremLet r be any positive number. Then
limx→∞
ex
xr= ∞.
Proof.If r is a positive integer, then apply L’Hôpital’s rule r times to thefraction. You get
limx→∞
ex
xrH= . . .
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, soex
xr>
ex
xm. The right-hand side tends to ∞ by the
previous step.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 24 / 34
. . . . . .
Exponential versus any power
TheoremLet r be any positive number. Then
limx→∞
ex
xr= ∞.
Proof.If r is a positive integer, then apply L’Hôpital’s rule r times to thefraction. You get
limx→∞
ex
xrH= . . .
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, soex
xr>
ex
xm. The right-hand side tends to ∞ by the
previous step.V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 24 / 34
. . . . . .
Any exponential versus any power
TheoremLet a > 1 and r > 0. Then
limx→∞
ax
xr= ∞.
Proof.If r is a positive integer, we have
limx→∞
ax
xrH= . . .
H= lim
x→∞
(lna)rax
r!= ∞.
If r isn’t an integer, we can compare it as before.
So even limx→∞
(1.00000001)x
x100000000= ∞!
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 25 / 34
. . . . . .
Any exponential versus any power
TheoremLet a > 1 and r > 0. Then
limx→∞
ax
xr= ∞.
Proof.If r is a positive integer, we have
limx→∞
ax
xrH= . . .
H= lim
x→∞
(lna)rax
r!= ∞.
If r isn’t an integer, we can compare it as before.
So even limx→∞
(1.00000001)x
x100000000= ∞!
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 25 / 34
. . . . . .
Any exponential versus any power
TheoremLet a > 1 and r > 0. Then
limx→∞
ax
xr= ∞.
Proof.If r is a positive integer, we have
limx→∞
ax
xrH= . . .
H= lim
x→∞
(lna)rax
r!= ∞.
If r isn’t an integer, we can compare it as before.
So even limx→∞
(1.00000001)x
x100000000= ∞!
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 25 / 34
. . . . . .
Logarithmic versus power growth
TheoremLet r be any positive number. Then
limx→∞
ln xxr
= 0.
Proof.One application of L’Hôpital’s Rule here suffices:
limx→∞
ln xxr
H= lim
x→∞
1/xrxr−1 = lim
x→∞
1rxr
= 0.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 26 / 34
. . . . . .
Logarithmic versus power growth
TheoremLet r be any positive number. Then
limx→∞
ln xxr
= 0.
Proof.One application of L’Hôpital’s Rule here suffices:
limx→∞
ln xxr
H= lim
x→∞
1/xrxr−1 = lim
x→∞
1rxr
= 0.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 26 / 34
. . . . . .
Outline
L’Hôpital’s Rule
Relative Rates of Growth
Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 27 / 34
. . . . . .
Indeterminate products
Example
Findlim
x→0+
√x ln x
This limit is of the form 0 · (−∞).
SolutionJury-rig the expression to make an indeterminate quotient. Then applyL’Hôpital’s Rule:
limx→0+
√x ln x
= limx→0+
ln x1/
√x
H= lim
x→0+x−1
−12x−3/2
= limx→0+
−2√x = 0
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34
. . . . . .
Indeterminate products
Example
Findlim
x→0+
√x ln x
This limit is of the form 0 · (−∞).
SolutionJury-rig the expression to make an indeterminate quotient. Then applyL’Hôpital’s Rule:
limx→0+
√x ln x
= limx→0+
ln x1/
√x
H= lim
x→0+x−1
−12x−3/2
= limx→0+
−2√x = 0
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34
. . . . . .
Indeterminate products
Example
Findlim
x→0+
√x ln x
This limit is of the form 0 · (−∞).
SolutionJury-rig the expression to make an indeterminate quotient. Then applyL’Hôpital’s Rule:
limx→0+
√x ln x = lim
x→0+ln x1/
√x
H= lim
x→0+x−1
−12x−3/2
= limx→0+
−2√x = 0
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34
. . . . . .
Indeterminate products
Example
Findlim
x→0+
√x ln x
This limit is of the form 0 · (−∞).
SolutionJury-rig the expression to make an indeterminate quotient. Then applyL’Hôpital’s Rule:
limx→0+
√x ln x = lim
x→0+ln x1/
√x
H= lim
x→0+x−1
−12x−3/2
= limx→0+
−2√x = 0
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34
. . . . . .
Indeterminate products
Example
Findlim
x→0+
√x ln x
This limit is of the form 0 · (−∞).
SolutionJury-rig the expression to make an indeterminate quotient. Then applyL’Hôpital’s Rule:
limx→0+
√x ln x = lim
x→0+ln x1/
√x
H= lim
x→0+x−1
−12x−3/2
= limx→0+
−2√x
= 0
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34
. . . . . .
Indeterminate products
Example
Findlim
x→0+
√x ln x
This limit is of the form 0 · (−∞).
SolutionJury-rig the expression to make an indeterminate quotient. Then applyL’Hôpital’s Rule:
limx→0+
√x ln x = lim
x→0+ln x1/
√x
H= lim
x→0+x−1
−12x−3/2
= limx→0+
−2√x = 0
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 28 / 34
. . . . . .
Indeterminate differences
Example
limx→0+
(1x− cot 2x
)
This limit is of the form ∞−∞.
SolutionAgain, rig it to make an indeterminate quotient.
limx→0+
sin(2x)− x cos(2x)x sin(2x)
H= lim
x→0+cos(2x) + 2x sin(2x)2x cos(2x) + sin(2x)
= ∞
The limit is +∞ becuase the numerator tends to 1 while thedenominator tends to zero but remains positive.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 29 / 34
. . . . . .
Indeterminate differences
Example
limx→0+
(1x− cot 2x
)
This limit is of the form ∞−∞.
SolutionAgain, rig it to make an indeterminate quotient.
limx→0+
sin(2x)− x cos(2x)x sin(2x)
H= lim
x→0+cos(2x) + 2x sin(2x)2x cos(2x) + sin(2x)
= ∞
The limit is +∞ becuase the numerator tends to 1 while thedenominator tends to zero but remains positive.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 29 / 34
. . . . . .
Indeterminate differences
Example
limx→0+
(1x− cot 2x
)
This limit is of the form ∞−∞.
SolutionAgain, rig it to make an indeterminate quotient.
limx→0+
sin(2x)− x cos(2x)x sin(2x)
H= lim
x→0+cos(2x) + 2x sin(2x)2x cos(2x) + sin(2x)
= ∞
The limit is +∞ becuase the numerator tends to 1 while thedenominator tends to zero but remains positive.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 29 / 34
. . . . . .
Indeterminate differences
Example
limx→0+
(1x− cot 2x
)
This limit is of the form ∞−∞.
SolutionAgain, rig it to make an indeterminate quotient.
limx→0+
sin(2x)− x cos(2x)x sin(2x)
H= lim
x→0+cos(2x) + 2x sin(2x)2x cos(2x) + sin(2x)
= ∞
The limit is +∞ becuase the numerator tends to 1 while thedenominator tends to zero but remains positive.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 29 / 34
. . . . . .
Indeterminate differences
Example
limx→0+
(1x− cot 2x
)
This limit is of the form ∞−∞.
SolutionAgain, rig it to make an indeterminate quotient.
limx→0+
sin(2x)− x cos(2x)x sin(2x)
H= lim
x→0+cos(2x) + 2x sin(2x)2x cos(2x) + sin(2x)
= ∞
The limit is +∞ becuase the numerator tends to 1 while thedenominator tends to zero but remains positive.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 29 / 34
. . . . . .
Checking your work
.
.limx→0
tan 2x2x
= 1, so for small x,
tan 2x ≈ 2x. So cot 2x ≈ 12x
and
1x− cot 2x ≈ 1
x− 1
2x=
12x
→ ∞
as x → 0+.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 30 / 34
. . . . . .
Indeterminate powers
Example
Find limx→0+
(1− 2x)1/x
Take the logarithm:
ln(
limx→0+
(1− 2x)1/x)
= limx→0+
ln((1− 2x)1/x
)= lim
x→0+ln(1− 2x)
x
This limit is of the form00, so we can use L’Hôpital:
limx→0+
ln(1− 2x)x
H= lim
x→0+
−21−2x1
= −2
This is not the answer, it’s the log of the answer! So the answer wewant is e−2.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34
. . . . . .
Indeterminate powers
Example
Find limx→0+
(1− 2x)1/x
Take the logarithm:
ln(
limx→0+
(1− 2x)1/x)
= limx→0+
ln((1− 2x)1/x
)= lim
x→0+ln(1− 2x)
x
This limit is of the form00, so we can use L’Hôpital:
limx→0+
ln(1− 2x)x
H= lim
x→0+
−21−2x1
= −2
This is not the answer, it’s the log of the answer! So the answer wewant is e−2.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34
. . . . . .
Indeterminate powers
Example
Find limx→0+
(1− 2x)1/x
Take the logarithm:
ln(
limx→0+
(1− 2x)1/x)
= limx→0+
ln((1− 2x)1/x
)= lim
x→0+ln(1− 2x)
x
This limit is of the form00, so we can use L’Hôpital:
limx→0+
ln(1− 2x)x
H= lim
x→0+
−21−2x1
= −2
This is not the answer, it’s the log of the answer! So the answer wewant is e−2.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34
. . . . . .
Indeterminate powers
Example
Find limx→0+
(1− 2x)1/x
Take the logarithm:
ln(
limx→0+
(1− 2x)1/x)
= limx→0+
ln((1− 2x)1/x
)= lim
x→0+ln(1− 2x)
x
This limit is of the form00, so we can use L’Hôpital:
limx→0+
ln(1− 2x)x
H= lim
x→0+
−21−2x1
= −2
This is not the answer, it’s the log of the answer! So the answer wewant is e−2.V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 31 / 34
. . . . . .
Another indeterminate power limit
Example
limx→0
(3x)4x
Solution
ln limx→0+
(3x)4x = limx→0+
ln(3x)4x = limx→0+
4x ln(3x)
= limx→0+
ln(3x)1/4x
H= lim
x→0+
3/3x−1/4x2
= limx→0+
(−4x) = 0
So the answer is e0 = 1.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 32 / 34
. . . . . .
Another indeterminate power limit
Example
limx→0
(3x)4x
Solution
ln limx→0+
(3x)4x = limx→0+
ln(3x)4x = limx→0+
4x ln(3x)
= limx→0+
ln(3x)1/4x
H= lim
x→0+
3/3x−1/4x2
= limx→0+
(−4x) = 0
So the answer is e0 = 1.
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 32 / 34
. . . . . .
Summary
Form Method00 L’Hôpital’s rule directly∞∞ L’Hôpital’s rule directly
0 · ∞ jiggle to make 00 or ∞
∞ .∞−∞ combine to make an indeterminate product or quotient
00 take ln to make an indeterminate product∞0 ditto1∞ ditto
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 33 / 34
. . . . . .
Final Thoughts
I L’Hôpital’s Rule only works on indeterminate quotientsI Luckily, most indeterminate limits can be transformed into
indeterminate quotientsI L’Hôpital’s Rule gives wrong answers for non-indeterminate limits!
V63.0121.041, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 3, 2010 34 / 34