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Section 3.7
    Indeterminate Forms and L’Hˆpital’s
                               o
                   Rule

                V63.0121.002.2010Su, Calculus I

                        New York University


                         June 7, 2010



Announcements
Announcements




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   2 / 26
Objectives



           Know when a limit is of
           indeterminate form:
                   indeterminate        quotients:
                   0/0, ∞/∞
                   indeterminate        products:
                   0×∞
                   indeterminate        differences:
                   ∞−∞
                   indeterminate        powers: 00 ,
                   ∞0 , and 1∞




V63.0121.002.2010Su, Calculus I (NYU)            L’Hˆpital’s Rule
                                                    o               June 7, 2010   3 / 26
Experiments with funny limits



              sin2 x
          lim
          x→0   x




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   4 / 26

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Experiments with funny limits



              sin2 x
          lim        =0
          x→0   x




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   4 / 26
Experiments with funny limits



              sin2 x
          lim        =0
          x→0   x
                x
          lim
          x→0 sin2 x




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   4 / 26
Experiments with funny limits



              sin2 x
          lim        =0
          x→0   x
                x
          lim        does not exist
          x→0 sin2 x




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   4 / 26
Experiments with funny limits



              sin2 x
          lim         =0
          x→0    x
                 x
          lim          does not exist
          x→0 sin2 x
               sin2 x
          lim
          x→0 sin(x 2 )




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   4 / 26

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Experiments with funny limits



              sin2 x
          lim         =0
          x→0    x
                 x
          lim          does not exist
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               sin2 x
          lim           =1
          x→0 sin(x 2 )




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   4 / 26
Experiments with funny limits



              sin2 x
          lim         =0
          x→0    x
                 x
          lim          does not exist
          x→0 sin2 x
               sin2 x
          lim           =1
          x→0 sin(x 2 )
              sin 3x
          lim
          x→0 sin x




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   4 / 26
Experiments with funny limits



              sin2 x
          lim         =0
          x→0    x
                 x
          lim          does not exist
          x→0 sin2 x
               sin2 x
          lim           =1
          x→0 sin(x 2 )
              sin 3x
          lim          =3
          x→0 sin x




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   4 / 26
Experiments with funny limits



            sin2 x
          lim       =0
       x→0     x
               x
       lim          does not exist
       x→0 sin2 x
             sin2 x
       lim           =1
       x→0 sin(x 2 )
            sin 3x
       lim          =3
       x→0 sin x
                                0
 All of these are of the form , and since we can get different answers in
                                0
 different cases, we say this form is indeterminate.




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o                 June 7, 2010   4 / 26

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Recall




 Recall the limit laws from Chapter 2.
         Limit of a sum is the sum of the limits




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   5 / 26
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




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o                   June 7, 2010   5 / 26
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




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o                   June 7, 2010   5 / 26
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.




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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:
                                          1                        cos x
                                   lim+     = +∞             lim         = −∞
                                 x→0      x                x→0−     x3




V63.0121.002.2010Su, Calculus I (NYU)         L’Hˆpital’s Rule
                                                 o                              June 7, 2010   6 / 26
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:
                                          1                            cos x
                                   lim+     = +∞                 lim         = −∞
                                 x→0      x                    x→0−     x3


         An exception would be something like
                                                    1
                                          lim            = lim x csc x.
                                          x→∞ 1    sin x  x→∞
                                              x

         which doesn’t exist.


V63.0121.002.2010Su, Calculus I (NYU)             L’Hˆpital’s Rule
                                                     o                              June 7, 2010   6 / 26
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:
                                          1                            cos x
                                   lim+     = +∞                 lim         = −∞
                                 x→0      x                    x→0−     x3


         An exception would be something like
                                                    1
                                          lim            = lim x csc x.
                                          x→∞ 1    sin x  x→∞
                                              x

         which doesn’t exist.
         Even less predictable: numerator and denominator both go to zero.

V63.0121.002.2010Su, Calculus I (NYU)             L’Hˆpital’s Rule
                                                     o                              June 7, 2010   6 / 26
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
            , and therefore nonexistent
          0
          because this expression is
          undefined.
                                    0
          The limit is of the form ,
                                    0
          which means we cannot
          evaluate it with our limit
          laws.



V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   7 / 26

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Indeterminate forms are like Tug Of War




 Which side wins depends on which side is stronger.

V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   8 / 26
Outline



 L’Hˆpital’s Rule
    o


 Other Indeterminate Limits
    Indeterminate Products
    Indeterminate Differences
    Indeterminate Powers


 Summary




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   9 / 26
The Linear Case
 Question
 If f and g are lines and f (a) = g (a) = 0, what is

                                            f (x)
                                        lim       ?
                                        x→a g (x)




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   10 / 26
The Linear Case
 Question
 If f and g are lines and f (a) = g (a) = 0, what is

                                               f (x)
                                            lim      ?
                                           x→a 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)   m1
                                                 =
                                           g (x)   m2

V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o               June 7, 2010   10 / 26

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The Linear Case, Illustrated


                            y                                  y = g (x)

                                                               y = f (x)


                                                       g (x)
                                        a   f (x)
                                                               x
                                                        x



                  f (x)   f (x) − f (a)   (f (x) − f (a))/(x − a)   m1
                        =               =                         =
                  g (x)   g (x) − g (a)   (g (x) − g (a))/(x − a)   m2


V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o                           June 7, 2010   11 / 26
What then?




         But what if the functions aren’t linear?




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   12 / 26
What then?




         But what if the functions aren’t linear?
         Can we approximate a function near a point with a linear function?




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o                   June 7, 2010   12 / 26
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?




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
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v6301210012011spv6301212011spv6301210112011sp
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!




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o                    June 7, 2010   12 / 26
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 f (x) = 0                      and             lim g (x) = 0
                x→a                                                x→a

 or

                 lim f (x) = ±∞                     and             lim g (x) = ±∞
                x→a                                                x→a

 Then
                                              f (x)       f (x)
                                        lim         = lim       ,
                                        x→a   g (x)   x→a g (x)

 if the limit on the right-hand side is finite, ∞, or −∞.

V63.0121.002.2010Su, Calculus I (NYU)           L’Hˆpital’s Rule
                                                   o                           June 7, 2010   13 / 26
Meet the Mathematician: L’Hˆpital
                           o



          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¸ois Antoine,
                                                                         c
                                                              Marquis de L’Hˆpital
                                                                             o
                                                              (French, 1661–1704)

V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o                               June 7, 2010   14 / 26
Revisiting the previous examples
 Example

                                     sin2 x
                                 lim
                                 x→0   x




V63.0121.002.2010Su, Calculus I (NYU)         L’Hˆpital’s Rule
                                                 o               June 7, 2010   15 / 26

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This document provides an overview of L'Hopital's rule for evaluating limits of indeterminate forms. It begins by defining different types of indeterminate forms such as 0/0, infinity/infinity, 0*infinity, infinity - infinity, and 00. It then introduces L'Hopital's rule, which allows such limits to be evaluated by taking the derivative of the numerator and denominator. Several examples are worked out to demonstrate how L'Hopital's rule can be applied. The document concludes by discussing various types of relative growth rates between functions as x approaches infinity.

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Revisiting the previous examples
 Example

                                     sin2 x H     2 sin x cos x
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V63.0121.002.2010Su, Calculus I (NYU)        L’Hˆpital’s Rule
                                                o                 June 7, 2010   15 / 26
Revisiting the previous examples
 Example                                                        sin x → 0

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V63.0121.002.2010Su, Calculus I (NYU)        L’Hˆpital’s Rule
                                                o                           June 7, 2010   15 / 26
Revisiting the previous examples
 Example                                                       sin x → 0

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                                 lim        = lim               =0
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V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o                           June 7, 2010   15 / 26
Revisiting the previous examples
 Example                                                       sin x → 0

                                     sin2 x H     2 sin x cos x
                                 lim        = lim               =0
                                 x→0   x      x→0       1


 Example

              sin2 x
           lim
          x→0 sin x 2




V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o                           June 7, 2010   15 / 26

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The document is a lecture on inverse trigonometric functions from a Calculus I class at New York University. It defines inverse trigonometric functions as the inverses of restricted trigonometric functions, gives their domains and ranges, and discusses their derivatives. The document also provides examples of evaluating inverse trigonometric functions.

calculusinverse trigonometricfunction
Lesson 12: Linear Approximation
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This document summarizes a calculus lecture on linear approximations. It provides examples of using the tangent line to approximate the sine function at different points. Specifically, it estimates sin(61°) by taking linear approximations about 0 and about 60°. The linear approximation about 0 is x, giving a value of 1.06465. The linear approximation about 60° uses the fact that the sine is √3/2 and the derivative is √3/2 at π/3, giving a better approximation than using 0.

calculusfunctionv6301210162010sp
Revisiting the previous examples
 Example

                                     sin2 x H     2 sin x cos x
                                 lim        = lim               =0
                                 x→0   x      x→0       1


 Example                    numerator → 0

              sin2 x
           lim
          x→0 sin x 2




V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o                     June 7, 2010   15 / 26
Revisiting the previous examples
 Example

                                     sin2 x H     2 sin x cos x
                                 lim        = lim               =0
                                 x→0   x      x→0       1


 Example                    numerator → 0

              sin2 x
           lim
          x→0 sin x 2




                          denominator → 0



V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o                     June 7, 2010   15 / 26
Revisiting the previous examples
 Example

                                     sin2 x H     2 sin x cos x
                                 lim        = lim               =0
                                 x→0   x      x→0       1


 Example                    numerator → 0

              sin2 x H       2 sin x cos x
                             ¡
           lim      2
                      = lim
          x→0 sin x     x→0 (cos x 2 ) (2x )
                                        ¡




                          denominator → 0



V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o                     June 7, 2010   15 / 26
Revisiting the previous examples
 Example

                                     sin2 x H     2 sin x cos x
                                 lim        = lim               =0
                                 x→0   x      x→0       1


 Example                                            numerator → 0

              sin2 x H       2 sin x cos x
                             ¡
           lim      2
                      = lim
          x→0 sin x     x→0 (cos x 2 ) (2x )
                                        ¡




V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o                     June 7, 2010   15 / 26

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This document is a section from a Calculus I course at New York University covering maximum and minimum values. It begins with announcements about exams and assignments. The objectives are to understand the Extreme Value Theorem and Fermat's Theorem, and use the Closed Interval Method to find extreme values. The document then covers the definitions of extreme points/values and the statements of the Extreme Value Theorem and Fermat's Theorem. It provides examples to illustrate the necessity of the hypotheses in the theorems. The focus is on using calculus concepts like continuity and differentiability to determine maximum and minimum values of functions on closed intervals.

Revisiting the previous examples
 Example

                                     sin2 x H     2 sin x cos x
                                 lim        = lim               =0
                                 x→0   x      x→0       1


 Example                                            numerator → 0

              sin2 x H       2 sin x cos x
                             ¡
           lim      2
                      = lim
          x→0 sin x     x→0 (cos x 2 ) (2x )
                                        ¡




                                                 denominator → 0



V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o                     June 7, 2010   15 / 26
Revisiting the previous examples
 Example

                                     sin2 x H     2 sin x cos x
                                 lim        = lim               =0
                                 x→0   x      x→0       1


 Example                                            numerator → 0

              sin2 x H       2 sin x cos x H
                             ¡                       cos2 x − sin2 x
           lim        = lim                  = lim
          x→0 sin x 2   x→0 (cos x 2 ) (2x )
                                        ¡      x→0 cos x 2 − 2x 2 sin(x 2 )




                                                 denominator → 0



V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o                       June 7, 2010   15 / 26
Revisiting the previous examples
 Example

                                     sin2 x H     2 sin x cos x
                                 lim        = lim               =0
                                 x→0   x      x→0       1


 Example                                                             numerator → 1

              sin2 x H       2 sin x cos x H
                             ¡                       cos2 x − sin2 x
           lim        = lim                  = lim
          x→0 sin x 2   x→0 (cos x 2 ) (2x )
                                        ¡      x→0 cos x 2 − 2x 2 sin(x 2 )




V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o                        June 7, 2010   15 / 26
Revisiting the previous examples
 Example

                                     sin2 x H     2 sin x cos x
                                 lim        = lim               =0
                                 x→0   x      x→0       1


 Example                                                              numerator → 1

              sin2 x H       2 sin x cos x H
                             ¡                       cos2 x − sin2 x
           lim        = lim                  = lim
          x→0 sin x 2   x→0 (cos x 2 ) (2x )
                                        ¡      x→0 cos x 2 − 2x 2 sin(x 2 )




                                                                     denominator → 1



V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o                         June 7, 2010   15 / 26

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Revisiting the previous examples
 Example

                                     sin2 x H     2 sin x cos x
                                 lim        = lim               =0
                                 x→0   x      x→0       1


 Example

              sin2 x H       2 sin x cos x H
                             ¡                       cos2 x − sin2 x
           lim        = lim                  = lim                          =1
          x→0 sin x 2   x→0 (cos x 2 ) (2x )
                                        ¡      x→0 cos x 2 − 2x 2 sin(x 2 )




V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o                     June 7, 2010   15 / 26
Revisiting the previous examples
 Example

                                     sin2 x H     2 sin x cos x
                                 lim        = lim               =0
                                 x→0   x      x→0       1


 Example

              sin2 x H       2 sin x cos x H
                             ¡                       cos2 x − sin2 x
           lim        = lim                  = lim                          =1
          x→0 sin x 2   x→0 (cos x 2 ) (2x )
                                        ¡      x→0 cos x 2 − 2x 2 sin(x 2 )



 Example

                                       sin 3x H     3 cos 3x
                                    lim       = lim          = 3.
                                   x→0 sin x    x→0 cos x


V63.0121.002.2010Su, Calculus I (NYU)        L’Hˆpital’s Rule
                                                o                    June 7, 2010   15 / 26
Another Example



 Example
 Find
                                                 x
                                        lim
                                        x→0    cos x




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   16 / 26
Beware of Red Herrings



 Example
 Find
                                                 x
                                        lim
                                        x→0    cos x

 Solution
 The limit of the denominator is 1, not 0, so L’Hˆpital’s rule does not
                                                 o
 apply. The limit is 0.




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o                  June 7, 2010   16 / 26

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Outline



 L’Hˆpital’s Rule
    o


 Other Indeterminate Limits
    Indeterminate Products
    Indeterminate Differences
    Indeterminate Powers


 Summary




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   17 / 26
Indeterminate products

 Example
 Find                                          √
                                        lim+       x ln x
                                        x→0


 This limit is of the form 0 · (−∞).




V63.0121.002.2010Su, Calculus I (NYU)    L’Hˆpital’s Rule
                                            o               June 7, 2010   18 / 26
Indeterminate products

 Example
 Find                                                √
                                              lim+       x ln x
                                              x→0


 This limit is of the form 0 · (−∞).
 Solution
 Jury-rig the expression to make an indeterminate quotient. Then apply
 L’Hˆpital’s Rule:
    o

                                 √
                          lim        x ln x
                         x→0+




V63.0121.002.2010Su, Calculus I (NYU)          L’Hˆpital’s Rule
                                                  o               June 7, 2010   18 / 26
Indeterminate products

 Example
 Find                                                √
                                             lim+        x ln x
                                             x→0


 This limit is of the form 0 · (−∞).
 Solution
 Jury-rig the expression to make an indeterminate quotient. Then apply
 L’Hˆpital’s Rule:
    o

                                 √                   ln x
                          lim        x ln x = lim+   1/√x
                         x→0+                x→0




V63.0121.002.2010Su, Calculus I (NYU)         L’Hˆpital’s Rule
                                                 o                June 7, 2010   18 / 26

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Indeterminate products

 Example
 Find                                                √
                                             lim+        x ln x
                                             x→0


 This limit is of the form 0 · (−∞).
 Solution
 Jury-rig the expression to make an indeterminate quotient. Then apply
 L’Hˆpital’s Rule:
    o

                                 √                    ln x H       x −1
                          lim        x ln x = lim+      √ = lim+
                         x→0+                x→0     1/ x    x→0 − 1 x −3/2
                                                                   2




V63.0121.002.2010Su, Calculus I (NYU)         L’Hˆpital’s Rule
                                                 o                            June 7, 2010   18 / 26
Indeterminate products

 Example
 Find                                                √
                                             lim+        x ln x
                                             x→0


 This limit is of the form 0 · (−∞).
 Solution
 Jury-rig the expression to make an indeterminate quotient. Then apply
 L’Hˆpital’s Rule:
    o

                                 √               ln x H       x −1
                          lim        x ln x = lim+ √ = lim+
                         x→0+              x→0 1/ x     x→0 − 1 x −3/2
                                                              2
                                                    √
                                          = lim+ −2 x
                                             x→0



V63.0121.002.2010Su, Calculus I (NYU)         L’Hˆpital’s Rule
                                                 o                       June 7, 2010   18 / 26
Indeterminate products

 Example
 Find                                                √
                                             lim+        x ln x
                                             x→0


 This limit is of the form 0 · (−∞).
 Solution
 Jury-rig the expression to make an indeterminate quotient. Then apply
 L’Hˆpital’s Rule:
    o

                                 √               ln x H       x −1
                          lim        x ln x = lim+ √ = lim+
                         x→0+              x→0 1/ x     x→0 − 1 x −3/2
                                                              2
                                                    √
                                          = lim+ −2 x = 0
                                             x→0



V63.0121.002.2010Su, Calculus I (NYU)         L’Hˆpital’s Rule
                                                 o                       June 7, 2010   18 / 26
Indeterminate differences

 Example

                                                  1
                                        lim+        − cot 2x
                                        x→0       x

 This limit is of the form ∞ − ∞.




V63.0121.002.2010Su, Calculus I (NYU)          L’Hˆpital’s Rule
                                                  o               June 7, 2010   19 / 26

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Indeterminate differences

 Example

                                                  1
                                        lim+        − cot 2x
                                        x→0       x

 This limit is of the form ∞ − ∞.
 Solution
 Again, rig it to make an indeterminate quotient.

                         sin(2x) − x cos(2x)
                  lim
                 x→0+         x sin(2x)




V63.0121.002.2010Su, Calculus I (NYU)          L’Hˆpital’s Rule
                                                  o               June 7, 2010   19 / 26
Indeterminate differences

 Example

                                                  1
                                        lim+        − cot 2x
                                        x→0       x

 This limit is of the form ∞ − ∞.
 Solution
 Again, rig it to make an indeterminate quotient.

                         sin(2x) − x cos(2x) H      cos(2x) + 2x sin(2x)
                  lim                        = lim+
                 x→0+         x sin(2x)        x→0 2x cos(2x) + sin(2x)




V63.0121.002.2010Su, Calculus I (NYU)          L’Hˆpital’s Rule
                                                  o                 June 7, 2010   19 / 26
Indeterminate differences

 Example

                                                  1
                                        lim+        − cot 2x
                                        x→0       x

 This limit is of the form ∞ − ∞.
 Solution
 Again, rig it to make an indeterminate quotient.

                         sin(2x) − x cos(2x) H      cos(2x) + 2x sin(2x)
                  lim                        = lim+
                 x→0+         x sin(2x)        x→0 2x cos(2x) + sin(2x)

                                             =∞




V63.0121.002.2010Su, Calculus I (NYU)          L’Hˆpital’s Rule
                                                  o                 June 7, 2010   19 / 26
Indeterminate differences

 Example

                                                  1
                                        lim+        − cot 2x
                                        x→0       x

 This limit is of the form ∞ − ∞.
 Solution
 Again, rig it to make an indeterminate quotient.

                         sin(2x) − x cos(2x) H      cos(2x) + 2x sin(2x)
                  lim                        = lim+
                 x→0+         x sin(2x)        x→0 2x cos(2x) + sin(2x)

                                             =∞

 The limit is +∞ becuase the numerator tends to 1 while the denominator
 tends to zero but remains positive.
V63.0121.002.2010Su, Calculus I (NYU)          L’Hˆpital’s Rule
                                                  o                 June 7, 2010   19 / 26

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                                   tan 2x
                               lim        = 1, so for small x,
                              x→0 2x
                                                         1
                              tan 2x ≈ 2x. So cot 2x ≈      and
                                                        2x
                                 1           1   1    1
                                   − cot 2x ≈ −    =    →∞
                                 x           x  2x   2x
                              as x → 0+ .




V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o                  June 7, 2010   20 / 26
Indeterminate powers

 Example
 Find lim+ (1 − 2x)1/x
         x→0




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   21 / 26
Indeterminate powers

 Example
 Find lim+ (1 − 2x)1/x
         x→0

 Take the logarithm:

                                                                       ln(1 − 2x)
       ln      lim+ (1 − 2x)1/x         = lim+ ln (1 − 2x)1/x = lim+
              x→0                         x→0                   x→0         x




V63.0121.002.2010Su, Calculus I (NYU)        L’Hˆpital’s Rule
                                                o                      June 7, 2010   21 / 26
Indeterminate powers

 Example
 Find lim+ (1 − 2x)1/x
         x→0

 Take the logarithm:

                                                                                  ln(1 − 2x)
       ln      lim+ (1 − 2x)1/x          = lim+ ln (1 − 2x)1/x = lim+
              x→0                           x→0                             x→0        x

                                        0
 This limit is of the form                , so we can use L’Hˆpital:
                                                             o
                                        0
                                                                   −2
                                        ln(1 − 2x) H              1−2x
                                lim+               = lim+                = −2
                               x→0           x       x→0           1




V63.0121.002.2010Su, Calculus I (NYU)          L’Hˆpital’s Rule
                                                  o                               June 7, 2010   21 / 26

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(msvu毕业证书)圣文森山大学毕业证
Indeterminate powers

 Example
 Find lim+ (1 − 2x)1/x
         x→0

 Take the logarithm:

                                                                                  ln(1 − 2x)
       ln      lim+ (1 − 2x)1/x          = lim+ ln (1 − 2x)1/x = lim+
              x→0                           x→0                             x→0        x

                                        0
 This limit is of the form                , so we can use L’Hˆpital:
                                                             o
                                        0
                                                                   −2
                                        ln(1 − 2x) H              1−2x
                                lim+               = lim+                = −2
                               x→0           x       x→0           1
 This is not the answer, it’s the log of the answer! So the answer we want
 is e −2 .
V63.0121.002.2010Su, Calculus I (NYU)          L’Hˆpital’s Rule
                                                  o                               June 7, 2010   21 / 26
Another indeterminate power limit

 Example

                                        lim (3x)4x
                                        x→0




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o               June 7, 2010   22 / 26
Another indeterminate power limit

 Example

                                            lim (3x)4x
                                            x→0



 Solution


                      ln lim+ (3x)4x = lim+ ln(3x)4x = lim+ 4x ln(3x)
                         x→0              x→0                  x→0
                                                ln(3x) H              3/3x
                                        = lim+    1/4x
                                                       = lim+        −1/4x 2
                                         x→0             x→0

                                        = lim+ (−4x) = 0
                                          x→0

 So the answer is e 0 = 1.

V63.0121.002.2010Su, Calculus I (NYU)       L’Hˆpital’s Rule
                                               o                               June 7, 2010   22 / 26
Summary
     Form          Method

         0
         0         L’Hˆpital’s rule directly
                      o

        ∞
        ∞          L’Hˆpital’s rule directly
                      o

                                        0        ∞
     0·∞           jiggle to make       0   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’Hˆpital’s Rule
                                                    o               June 7, 2010   23 / 26

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Final Thoughts




         L’Hˆpital’s Rule only works on indeterminate quotients
            o
         Luckily, most indeterminate limits can be transformed into
         indeterminate quotients
         L’Hˆpital’s Rule gives wrong answers for non-indeterminate limits!
            o




V63.0121.002.2010Su, Calculus I (NYU)   L’Hˆpital’s Rule
                                           o                      June 7, 2010   24 / 26

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Lesson 17: Indeterminate Forms and L'Hôpital's Rule

  • 1. Section 3.7 Indeterminate Forms and L’Hˆpital’s o Rule V63.0121.002.2010Su, Calculus I New York University June 7, 2010 Announcements
  • 2. Announcements V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 2 / 26
  • 3. Objectives Know when a limit is of indeterminate form: indeterminate quotients: 0/0, ∞/∞ indeterminate products: 0×∞ indeterminate differences: ∞−∞ indeterminate powers: 00 , ∞0 , and 1∞ V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 3 / 26
  • 4. Experiments with funny limits sin2 x lim x→0 x V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 4 / 26
  • 5. Experiments with funny limits sin2 x lim =0 x→0 x V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 4 / 26
  • 6. Experiments with funny limits sin2 x lim =0 x→0 x x lim x→0 sin2 x V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 4 / 26
  • 7. Experiments with funny limits sin2 x lim =0 x→0 x x lim does not exist x→0 sin2 x V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 4 / 26
  • 8. Experiments with funny limits sin2 x lim =0 x→0 x x lim does not exist x→0 sin2 x sin2 x lim x→0 sin(x 2 ) V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 4 / 26
  • 9. Experiments with funny limits sin2 x lim =0 x→0 x x lim does not exist x→0 sin2 x sin2 x lim =1 x→0 sin(x 2 ) V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 4 / 26
  • 10. Experiments with funny limits sin2 x lim =0 x→0 x x lim does not exist x→0 sin2 x sin2 x lim =1 x→0 sin(x 2 ) sin 3x lim x→0 sin x V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 4 / 26
  • 11. Experiments with funny limits sin2 x lim =0 x→0 x x lim does not exist x→0 sin2 x sin2 x lim =1 x→0 sin(x 2 ) sin 3x lim =3 x→0 sin x V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 4 / 26
  • 12. Experiments with funny limits sin2 x lim =0 x→0 x x lim does not exist x→0 sin2 x sin2 x lim =1 x→0 sin(x 2 ) sin 3x lim =3 x→0 sin x 0 All of these are of the form , and since we can get different answers in 0 different cases, we say this form is indeterminate. V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 4 / 26
  • 13. Recall Recall the limit laws from Chapter 2. Limit of a sum is the sum of the limits V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 5 / 26
  • 14. 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 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 5 / 26
  • 15. 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 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 5 / 26
  • 16. 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. V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 5 / 26
  • 17. 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: 1 cos x lim+ = +∞ lim = −∞ x→0 x x→0− x3 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 6 / 26
  • 18. 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: 1 cos x lim+ = +∞ lim = −∞ x→0 x x→0− x3 An exception would be something like 1 lim = lim x csc x. x→∞ 1 sin x x→∞ x which doesn’t exist. V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 6 / 26
  • 19. 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: 1 cos x lim+ = +∞ lim = −∞ x→0 x x→0− x3 An exception would be something like 1 lim = lim x csc x. x→∞ 1 sin x x→∞ x which doesn’t exist. Even less predictable: numerator and denominator both go to zero. V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 6 / 26
  • 20. 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 , and therefore nonexistent 0 because this expression is undefined. 0 The limit is of the form , 0 which means we cannot evaluate it with our limit laws. V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 7 / 26
  • 21. Indeterminate forms are like Tug Of War Which side wins depends on which side is stronger. V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 8 / 26
  • 22. Outline L’Hˆpital’s Rule o Other Indeterminate Limits Indeterminate Products Indeterminate Differences Indeterminate Powers Summary V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 9 / 26
  • 23. The Linear Case Question If f and g are lines and f (a) = g (a) = 0, what is f (x) lim ? x→a g (x) V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 10 / 26
  • 24. The Linear Case Question If f and g are lines and f (a) = g (a) = 0, what is f (x) lim ? x→a 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) m1 = g (x) m2 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 10 / 26
  • 25. The Linear Case, Illustrated y y = g (x) y = f (x) g (x) a f (x) x x f (x) f (x) − f (a) (f (x) − f (a))/(x − a) m1 = = = g (x) g (x) − g (a) (g (x) − g (a))/(x − a) m2 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 11 / 26
  • 26. What then? But what if the functions aren’t linear? V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 12 / 26
  • 27. What then? But what if the functions aren’t linear? Can we approximate a function near a point with a linear function? V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 12 / 26
  • 28. 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? V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 12 / 26
  • 29. 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! V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 12 / 26
  • 30. 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 f (x) = 0 and lim g (x) = 0 x→a x→a or lim f (x) = ±∞ and lim g (x) = ±∞ x→a x→a Then f (x) f (x) lim = lim , x→a g (x) x→a g (x) if the limit on the right-hand side is finite, ∞, or −∞. V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 13 / 26
  • 31. Meet the Mathematician: L’Hˆpital o 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¸ois Antoine, c Marquis de L’Hˆpital o (French, 1661–1704) V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 14 / 26
  • 32. Revisiting the previous examples Example sin2 x lim x→0 x V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 33. Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim x→0 x x→0 1 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 34. Revisiting the previous examples Example sin x → 0 sin2 x H 2 sin x cos x lim = lim x→0 x x→0 1 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 35. Revisiting the previous examples Example sin x → 0 sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 36. Revisiting the previous examples Example sin x → 0 sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example sin2 x lim x→0 sin x 2 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 37. Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example numerator → 0 sin2 x lim x→0 sin x 2 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 38. Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example numerator → 0 sin2 x lim x→0 sin x 2 denominator → 0 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 39. Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example numerator → 0 sin2 x H 2 sin x cos x ¡ lim 2 = lim x→0 sin x x→0 (cos x 2 ) (2x ) ¡ denominator → 0 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 40. Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example numerator → 0 sin2 x H 2 sin x cos x ¡ lim 2 = lim x→0 sin x x→0 (cos x 2 ) (2x ) ¡ V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 41. Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example numerator → 0 sin2 x H 2 sin x cos x ¡ lim 2 = lim x→0 sin x x→0 (cos x 2 ) (2x ) ¡ denominator → 0 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 42. Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example numerator → 0 sin2 x H 2 sin x cos x H ¡ cos2 x − sin2 x lim = lim = lim x→0 sin x 2 x→0 (cos x 2 ) (2x ) ¡ x→0 cos x 2 − 2x 2 sin(x 2 ) denominator → 0 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 43. Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example numerator → 1 sin2 x H 2 sin x cos x H ¡ cos2 x − sin2 x lim = lim = lim x→0 sin x 2 x→0 (cos x 2 ) (2x ) ¡ x→0 cos x 2 − 2x 2 sin(x 2 ) V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 44. Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example numerator → 1 sin2 x H 2 sin x cos x H ¡ cos2 x − sin2 x lim = lim = lim x→0 sin x 2 x→0 (cos x 2 ) (2x ) ¡ x→0 cos x 2 − 2x 2 sin(x 2 ) denominator → 1 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 45. Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example sin2 x H 2 sin x cos x H ¡ cos2 x − sin2 x lim = lim = lim =1 x→0 sin x 2 x→0 (cos x 2 ) (2x ) ¡ x→0 cos x 2 − 2x 2 sin(x 2 ) V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 46. Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim =0 x→0 x x→0 1 Example sin2 x H 2 sin x cos x H ¡ cos2 x − sin2 x lim = lim = lim =1 x→0 sin x 2 x→0 (cos x 2 ) (2x ) ¡ x→0 cos x 2 − 2x 2 sin(x 2 ) Example sin 3x H 3 cos 3x lim = lim = 3. x→0 sin x x→0 cos x V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 15 / 26
  • 47. Another Example Example Find x lim x→0 cos x V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 16 / 26
  • 48. Beware of Red Herrings Example Find x lim x→0 cos x Solution The limit of the denominator is 1, not 0, so L’Hˆpital’s rule does not o apply. The limit is 0. V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 16 / 26
  • 49. Outline L’Hˆpital’s Rule o Other Indeterminate Limits Indeterminate Products Indeterminate Differences Indeterminate Powers Summary V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 17 / 26
  • 50. Indeterminate products Example Find √ lim+ x ln x x→0 This limit is of the form 0 · (−∞). V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 18 / 26
  • 51. Indeterminate products Example Find √ lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solution Jury-rig the expression to make an indeterminate quotient. Then apply L’Hˆpital’s Rule: o √ lim x ln x x→0+ V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 18 / 26
  • 52. Indeterminate products Example Find √ lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solution Jury-rig the expression to make an indeterminate quotient. Then apply L’Hˆpital’s Rule: o √ ln x lim x ln x = lim+ 1/√x x→0+ x→0 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 18 / 26
  • 53. Indeterminate products Example Find √ lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solution Jury-rig the expression to make an indeterminate quotient. Then apply L’Hˆpital’s Rule: o √ ln x H x −1 lim x ln x = lim+ √ = lim+ x→0+ x→0 1/ x x→0 − 1 x −3/2 2 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 18 / 26
  • 54. Indeterminate products Example Find √ lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solution Jury-rig the expression to make an indeterminate quotient. Then apply L’Hˆpital’s Rule: o √ ln x H x −1 lim x ln x = lim+ √ = lim+ x→0+ x→0 1/ x x→0 − 1 x −3/2 2 √ = lim+ −2 x x→0 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 18 / 26
  • 55. Indeterminate products Example Find √ lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solution Jury-rig the expression to make an indeterminate quotient. Then apply L’Hˆpital’s Rule: o √ ln x H x −1 lim x ln x = lim+ √ = lim+ x→0+ x→0 1/ x x→0 − 1 x −3/2 2 √ = lim+ −2 x = 0 x→0 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 18 / 26
  • 56. Indeterminate differences Example 1 lim+ − cot 2x x→0 x This limit is of the form ∞ − ∞. V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 19 / 26
  • 57. Indeterminate differences Example 1 lim+ − cot 2x x→0 x This limit is of the form ∞ − ∞. Solution Again, rig it to make an indeterminate quotient. sin(2x) − x cos(2x) lim x→0+ x sin(2x) V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 19 / 26
  • 58. Indeterminate differences Example 1 lim+ − cot 2x x→0 x This limit is of the form ∞ − ∞. Solution Again, rig it to make an indeterminate quotient. sin(2x) − x cos(2x) H cos(2x) + 2x sin(2x) lim = lim+ x→0+ x sin(2x) x→0 2x cos(2x) + sin(2x) V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 19 / 26
  • 59. Indeterminate differences Example 1 lim+ − cot 2x x→0 x This limit is of the form ∞ − ∞. Solution Again, rig it to make an indeterminate quotient. sin(2x) − x cos(2x) H cos(2x) + 2x sin(2x) lim = lim+ x→0+ x sin(2x) x→0 2x cos(2x) + sin(2x) =∞ V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 19 / 26
  • 60. Indeterminate differences Example 1 lim+ − cot 2x x→0 x This limit is of the form ∞ − ∞. Solution Again, rig it to make an indeterminate quotient. sin(2x) − x cos(2x) H cos(2x) + 2x sin(2x) lim = lim+ x→0+ x sin(2x) x→0 2x cos(2x) + sin(2x) =∞ The limit is +∞ becuase the numerator tends to 1 while the denominator tends to zero but remains positive. V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 19 / 26
  • 61. Checking your work tan 2x lim = 1, so for small x, x→0 2x 1 tan 2x ≈ 2x. So cot 2x ≈ and 2x 1 1 1 1 − cot 2x ≈ − = →∞ x x 2x 2x as x → 0+ . V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 20 / 26
  • 62. Indeterminate powers Example Find lim+ (1 − 2x)1/x x→0 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 21 / 26
  • 63. Indeterminate powers Example Find lim+ (1 − 2x)1/x x→0 Take the logarithm: ln(1 − 2x) ln lim+ (1 − 2x)1/x = lim+ ln (1 − 2x)1/x = lim+ x→0 x→0 x→0 x V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 21 / 26
  • 64. Indeterminate powers Example Find lim+ (1 − 2x)1/x x→0 Take the logarithm: ln(1 − 2x) ln lim+ (1 − 2x)1/x = lim+ ln (1 − 2x)1/x = lim+ x→0 x→0 x→0 x 0 This limit is of the form , so we can use L’Hˆpital: o 0 −2 ln(1 − 2x) H 1−2x lim+ = lim+ = −2 x→0 x x→0 1 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 21 / 26
  • 65. Indeterminate powers Example Find lim+ (1 − 2x)1/x x→0 Take the logarithm: ln(1 − 2x) ln lim+ (1 − 2x)1/x = lim+ ln (1 − 2x)1/x = lim+ x→0 x→0 x→0 x 0 This limit is of the form , so we can use L’Hˆpital: o 0 −2 ln(1 − 2x) H 1−2x lim+ = lim+ = −2 x→0 x x→0 1 This is not the answer, it’s the log of the answer! So the answer we want is e −2 . V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 21 / 26
  • 66. Another indeterminate power limit Example lim (3x)4x x→0 V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 22 / 26
  • 67. Another indeterminate power limit Example lim (3x)4x x→0 Solution ln lim+ (3x)4x = lim+ ln(3x)4x = lim+ 4x ln(3x) x→0 x→0 x→0 ln(3x) H 3/3x = lim+ 1/4x = lim+ −1/4x 2 x→0 x→0 = lim+ (−4x) = 0 x→0 So the answer is e 0 = 1. V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 22 / 26
  • 68. Summary Form Method 0 0 L’Hˆpital’s rule directly o ∞ ∞ L’Hˆpital’s rule directly o 0 ∞ 0·∞ jiggle to make 0 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’Hˆpital’s Rule o June 7, 2010 23 / 26
  • 69. Final Thoughts L’Hˆpital’s Rule only works on indeterminate quotients o Luckily, most indeterminate limits can be transformed into indeterminate quotients L’Hˆpital’s Rule gives wrong answers for non-indeterminate limits! o V63.0121.002.2010Su, Calculus I (NYU) L’Hˆpital’s Rule o June 7, 2010 24 / 26