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Section 3.7Indeterminate Forms and L’Hopital’s
Rule
V63.0121.002.2010Su, Calculus I
New York University
June 7, 2010
Announcements
I
Announcements
I
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 2 / 26
Objectives
I Know when a limit is ofindeterminate form:
I indeterminate quotients:0/0, ∞/∞
I indeterminate products:0×∞
I indeterminate differences:∞−∞
I indeterminate powers: 00,∞0, and 1∞
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 3 / 26
Experiments with funny limits
I limx→0
sin2 x
x
= 0
I limx→0
x
sin2 x
does not exist
I limx→0
sin2 x
sin(x2)
= 1
I limx→0
sin 3x
sin x
= 3
All of these are of the form0
0, and since we can get different answers in
different cases, we say this form is indeterminate.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 4 / 26
Experiments with funny limits
I limx→0
sin2 x
x= 0
I limx→0
x
sin2 x
does not exist
I limx→0
sin2 x
sin(x2)
= 1
I limx→0
sin 3x
sin x
= 3
All of these are of the form0
0, and since we can get different answers in
different cases, we say this form is indeterminate.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 4 / 26
Experiments with funny limits
I limx→0
sin2 x
x= 0
I limx→0
x
sin2 x
does not exist
I limx→0
sin2 x
sin(x2)
= 1
I limx→0
sin 3x
sin x
= 3
All of these are of the form0
0, and since we can get different answers in
different cases, we say this form is indeterminate.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 4 / 26
Experiments with funny limits
I limx→0
sin2 x
x= 0
I limx→0
x
sin2 xdoes not exist
I limx→0
sin2 x
sin(x2)
= 1
I limx→0
sin 3x
sin x
= 3
All of these are of the form0
0, and since we can get different answers in
different cases, we say this form is indeterminate.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 4 / 26
Experiments with funny limits
I limx→0
sin2 x
x= 0
I limx→0
x
sin2 xdoes not exist
I limx→0
sin2 x
sin(x2)
= 1
I limx→0
sin 3x
sin x
= 3
All of these are of the form0
0, and since we can get different answers in
different cases, we say this form is indeterminate.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 4 / 26
Experiments with funny limits
I limx→0
sin2 x
x= 0
I limx→0
x
sin2 xdoes not exist
I limx→0
sin2 x
sin(x2)= 1
I limx→0
sin 3x
sin x
= 3
All of these are of the form0
0, and since we can get different answers in
different cases, we say this form is indeterminate.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 4 / 26
Experiments with funny limits
I limx→0
sin2 x
x= 0
I limx→0
x
sin2 xdoes not exist
I limx→0
sin2 x
sin(x2)= 1
I limx→0
sin 3x
sin x
= 3
All of these are of the form0
0, and since we can get different answers in
different cases, we say this form is indeterminate.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 4 / 26
Experiments with funny limits
I limx→0
sin2 x
x= 0
I limx→0
x
sin2 xdoes not exist
I limx→0
sin2 x
sin(x2)= 1
I limx→0
sin 3x
sin x= 3
All of these are of the form0
0, and since we can get different answers in
different cases, we say this form is indeterminate.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 4 / 26
Experiments with funny limits
I limx→0
sin2 x
x= 0
I limx→0
x
sin2 xdoes not exist
I limx→0
sin2 x
sin(x2)= 1
I limx→0
sin 3x
sin x= 3
All of these are of the form0
0, and since we can get different answers in
different cases, we say this form is indeterminate.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 4 / 26
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 limits
I 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.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 5 / 26
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 limits
I 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.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 5 / 26
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 limits
I 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.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 5 / 26
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 limits
I 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.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 5 / 26
More about dividing limits
I We know dividing by zero is bad.
I Most of the time, if an expression’s numerator approaches a finitenumber and denominator approaches zero, the quotient approachessome kind of infinity. For example:
limx→0+
1
x= +∞ lim
x→0−
cos x
x3= −∞
I An exception would be something like
limx→∞
11x sin x
= limx→∞
x csc x .
which doesn’t exist.
I Even less predictable: numerator and denominator both go to zero.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 6 / 26
More about dividing limits
I We know dividing by zero is bad.
I Most of the time, if an expression’s numerator approaches a finitenumber and denominator approaches zero, the quotient approachessome kind of infinity. For example:
limx→0+
1
x= +∞ lim
x→0−
cos x
x3= −∞
I An exception would be something like
limx→∞
11x sin x
= limx→∞
x csc x .
which doesn’t exist.
I Even less predictable: numerator and denominator both go to zero.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 6 / 26
More about dividing limits
I We know dividing by zero is bad.
I Most of the time, if an expression’s numerator approaches a finitenumber and denominator approaches zero, the quotient approachessome kind of infinity. For example:
limx→0+
1
x= +∞ lim
x→0−
cos x
x3= −∞
I An exception would be something like
limx→∞
11x sin x
= limx→∞
x csc x .
which doesn’t exist.
I Even less predictable: numerator and denominator both go to zero.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 6 / 26
Language NoteIt depends on what the meaning of the word “is” is
I Be careful with the languagehere. We are not saying thatthe limit in each case “is”0
0, and therefore nonexistent
because this expression isundefined.
I The limit is of the form0
0,
which means we cannotevaluate it with our limitlaws.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 7 / 26
Indeterminate forms are like Tug Of War
Which side wins depends on which side is stronger.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 8 / 26
Outline
L’Hopital’s Rule
Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers
Summary
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 9 / 26
The Linear Case
Question
If f and g are lines and f (a) = g(a) = 0, what is
limx→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)
Sof (x)
g(x)=
m1
m2
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 10 / 26
The Linear Case
Question
If f and g are lines and f (a) = g(a) = 0, what is
limx→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)
Sof (x)
g(x)=
m1
m2
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 10 / 26
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
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 11 / 26
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!
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 12 / 26
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!
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 12 / 26
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!
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 12 / 26
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!
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 12 / 26
Theorem of the Day
Theorem (L’Hopital’s Rule)
Suppose f and g are differentiable functions and g ′(x) 6= 0 near a (exceptpossibly 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) = ±∞
Then
limx→a
f (x)
g(x)= lim
x→a
f ′(x)
g ′(x),
if the limit on the right-hand side is finite, ∞, or −∞.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 13 / 26
Meet the Mathematician: L’Hopital
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 Francois Antoine,
Marquis de L’Hopital(French, 1661–1704)
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 14 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
x
H= lim
x→0
2 sin x
sin x → 0
cos x
1= 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 3x
sin x
H= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x
sin x → 0
cos x
1
= 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 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x
sin x → 0
cos x
1
= 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 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x
sin x → 0
cos x
1= 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 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x
sin x → 0
cos x
1= 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 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x
sin x → 0
cos x
1= 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 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x
sin x → 0
cos x
1= 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 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x
sin x → 0
cos x
1= 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 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x
sin x → 0
cos x
1= 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 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x
sin x → 0
cos x
1= 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 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x
sin x → 0
cos x
1= 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 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x
sin x → 0
cos x
1= 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 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x
sin x → 0
cos x
1= 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 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x
sin x → 0
cos x
1= 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 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Revisiting the previous examples
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x
sin x → 0
cos x
1= 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 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 15 / 26
Another Example
Example
Findlimx→0
x
cos x
Solution
The limit of the denominator is 1, not 0, so L’Hopital’s rule does notapply. The limit is 0.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 16 / 26
Beware of Red Herrings
Example
Findlimx→0
x
cos x
Solution
The limit of the denominator is 1, not 0, so L’Hopital’s rule does notapply. The limit is 0.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 16 / 26
Outline
L’Hopital’s Rule
Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers
Summary
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 17 / 26
Indeterminate products
Example
Findlim
x→0+
√x ln x
This limit is of the form 0 · (−∞).
Solution
Jury-rig the expression to make an indeterminate quotient. Then applyL’Hopital’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
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 18 / 26
Indeterminate products
Example
Findlim
x→0+
√x ln x
This limit is of the form 0 · (−∞).
Solution
Jury-rig the expression to make an indeterminate quotient. Then applyL’Hopital’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
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 18 / 26
Indeterminate products
Example
Findlim
x→0+
√x ln x
This limit is of the form 0 · (−∞).
Solution
Jury-rig the expression to make an indeterminate quotient. Then applyL’Hopital’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
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 18 / 26
Indeterminate products
Example
Findlim
x→0+
√x ln x
This limit is of the form 0 · (−∞).
Solution
Jury-rig the expression to make an indeterminate quotient. Then applyL’Hopital’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
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 18 / 26
Indeterminate products
Example
Findlim
x→0+
√x ln x
This limit is of the form 0 · (−∞).
Solution
Jury-rig the expression to make an indeterminate quotient. Then applyL’Hopital’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
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 18 / 26
Indeterminate products
Example
Findlim
x→0+
√x ln x
This limit is of the form 0 · (−∞).
Solution
Jury-rig the expression to make an indeterminate quotient. Then applyL’Hopital’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
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 18 / 26
Indeterminate differences
Example
limx→0+
(1
x− cot 2x
)
This limit is of the form ∞−∞.
Solution
Again, 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 the denominatortends to zero but remains positive.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 19 / 26
Indeterminate differences
Example
limx→0+
(1
x− cot 2x
)
This limit is of the form ∞−∞.
Solution
Again, 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 the denominatortends to zero but remains positive.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 19 / 26
Indeterminate differences
Example
limx→0+
(1
x− cot 2x
)
This limit is of the form ∞−∞.
Solution
Again, 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 the denominatortends to zero but remains positive.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 19 / 26
Indeterminate differences
Example
limx→0+
(1
x− cot 2x
)
This limit is of the form ∞−∞.
Solution
Again, 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 the denominatortends to zero but remains positive.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 19 / 26
Indeterminate differences
Example
limx→0+
(1
x− cot 2x
)
This limit is of the form ∞−∞.
Solution
Again, 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 the denominatortends to zero but remains positive.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 19 / 26
Checking your work
limx→0
tan 2x
2x= 1, so for small x ,
tan 2x ≈ 2x . So cot 2x ≈ 1
2xand
1
x− cot 2x ≈ 1
x− 1
2x=
1
2x→∞
as x → 0+.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 20 / 26
Indeterminate powers
Example
Find limx→0+
(1− 2x)1/x
Take the logarithm:
ln
(lim
x→0+(1− 2x)1/x
)= lim
x→0+ln(
(1− 2x)1/x)
= limx→0+
ln(1− 2x)
x
This limit is of the form0
0, so we can use L’Hopital:
limx→0+
ln(1− 2x)
xH= lim
x→0+
−21−2x
1= −2
This is not the answer, it’s the log of the answer! So the answer we wantis e−2.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 21 / 26
Indeterminate powers
Example
Find limx→0+
(1− 2x)1/x
Take the logarithm:
ln
(lim
x→0+(1− 2x)1/x
)= lim
x→0+ln(
(1− 2x)1/x)
= limx→0+
ln(1− 2x)
x
This limit is of the form0
0, so we can use L’Hopital:
limx→0+
ln(1− 2x)
xH= lim
x→0+
−21−2x
1= −2
This is not the answer, it’s the log of the answer! So the answer we wantis e−2.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 21 / 26
Indeterminate powers
Example
Find limx→0+
(1− 2x)1/x
Take the logarithm:
ln
(lim
x→0+(1− 2x)1/x
)= lim
x→0+ln(
(1− 2x)1/x)
= limx→0+
ln(1− 2x)
x
This limit is of the form0
0, so we can use L’Hopital:
limx→0+
ln(1− 2x)
xH= lim
x→0+
−21−2x
1= −2
This is not the answer, it’s the log of the answer! So the answer we wantis e−2.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 21 / 26
Indeterminate powers
Example
Find limx→0+
(1− 2x)1/x
Take the logarithm:
ln
(lim
x→0+(1− 2x)1/x
)= lim
x→0+ln(
(1− 2x)1/x)
= limx→0+
ln(1− 2x)
x
This limit is of the form0
0, so we can use L’Hopital:
limx→0+
ln(1− 2x)
xH= lim
x→0+
−21−2x
1= −2
This is not the answer, it’s the log of the answer! So the answer we wantis e−2.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 21 / 26
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.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 22 / 26
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.
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 22 / 26
Summary
Form Method
00 L’Hopital’s rule directly
∞∞ L’Hopital’s rule directly
0 · ∞ jiggle to make 00 or ∞∞ .
∞−∞ factor to make an indeterminate product
00 take ln to make an indeterminate product
∞0 ditto
1∞ ditto
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 23 / 26
Final Thoughts
I L’Hopital’s Rule only works on indeterminate quotients
I Luckily, most indeterminate limits can be transformed intoindeterminate quotients
I L’Hopital’s Rule gives wrong answers for non-indeterminate limits!
V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 24 / 26