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Sec on 3.7
    Indeterminate forms and lHôpital’s
                  Rule
               V63.0121.011: Calculus I
             Professor Ma hew Leingang
                    New York University


                  March 30, 2011

.
Announcements

   Midterm has been
   returned. Please see FAQ
   on Blackboard (under
   ”Exams and Quizzes”)
   Quiz 3 this week in
   recita on on Sec on 2.6,
   2.8, 3.1, 3.2
Objectives
   Know when a limit is of
   indeterminate form:
       indeterminate quo ents:
       0/0, ∞/∞
       indeterminate products:
       0×∞
       indeterminate
       differences: ∞ − ∞
       indeterminate powers:
       00 , ∞0 , and 1∞
   Resolve limits in
   indeterminate form
Recall

 Recall the limit laws from Chapter 2.
     Limit of a sum is the sum of the limits
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
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
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 quo ent is the quo ent of the limits ... whoops! This
     is true as long as you don’t try to divide by zero.
More about dividing limits

   We know dividing by zero is bad.
   Most of the me, if an expression’s numerator approaches a
   finite nonzero number and denominator approaches zero, the
   quo ent has an infinite. For example:
                      1                 cos x
               lim+     = +∞     lim−         = −∞
               x→0    x         x→0      x3
Why 1/0 ̸= ∞
                                    1
 Consider the func on f(x) =   1         .
                               x   sin x
            y

                .
                                                           x


 Then lim f(x) is of the form 1/0, but the limit does not exist and is
       x→∞
 not infinite.
Why 1/0 ̸= ∞
                                    1
 Consider the func on f(x) =   1         .
                               x   sin x
           y

               .
                                                          x


 Then lim f(x) is of the form 1/0, but the limit does not exist and is
       x→∞
 not infinite.
 Even less predictable: when numerator and denominator both go to
 zero.
Experiments with funny limits
        sin2 x
    lim
    x→0   x

                         .
Experiments with funny limits
        sin2 x
    lim        =0
    x→0   x

                         .
Experiments with funny limits
        sin2 x
    lim        =0
    x→0   x
          x
    lim 2
    x→0 sin x
                         .
Experiments with funny limits
        sin2 x
    lim        =0
    x→0   x
          x
    lim 2 does not exist
    x→0 sin x
                           .
Experiments with funny limits
        sin2 x
    lim         =0
    x→0    x
           x
    lim 2 does not exist
    x→0 sin x
         sin2 x            .
    lim
    x→0 sin(x2 )
Experiments with funny limits
        sin2 x
    lim         =0
    x→0    x
           x
    lim 2 does not exist
    x→0 sin x
         sin2 x            .
    lim          =1
    x→0 sin(x2 )
Experiments with funny limits
        sin2 x
    lim         =0
    x→0    x
           x
    lim 2 does not exist
    x→0 sin x
         sin2 x            .
    lim          =1
    x→0 sin(x2 )
        sin 3x
    lim
    x→0 sin x
Experiments with funny limits
        sin2 x
    lim         =0
    x→0    x
           x
    lim 2 does not exist
    x→0 sin x
         sin2 x            .
    lim          =1
    x→0 sin(x2 )
        sin 3x
    lim         =3
    x→0 sin x
Experiments with funny limits
            sin2 x
        lim         =0
        x→0    x
               x
        lim 2 does not exist
        x→0 sin x
             sin2 x                                   .
        lim          =1
        x→0 sin(x2 )
            sin 3x
        lim         =3
        x→0 sin x
                             0
 All of these are of the form , and since we can get different
                             0
 answers in different cases, we say this form is indeterminate.
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
               0
     case “is” , and therefore nonexistent
               0
     because this expression is undefined.
                              0
     The limit is of the form , which means
                              0
     we cannot evaluate it with our limit
     laws.
Indeterminate forms are like Tug Of War




  Which side wins depends on which side is stronger.
Outline
 L’Hôpital’s Rule

 Rela ve Rates of Growth

 Other Indeterminate Limits
    Indeterminate Products
    Indeterminate Differences
    Indeterminate Powers
The Linear Case
 Ques on
                                                      f(x)
 If f and g are lines and f(a) = g(a) = 0, what is lim     ?
                                                  x→a g(x)
The Linear Case
 Ques on
                                                       f(x)
 If f and g are lines and f(a) = g(a) = 0, what is lim      ?
                                                   x→a g(x)


 Solu on
 The func ons f and g can be wri en in the form

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

 So
                                f(x)   m1
                                     =
                                g(x) m2
The Linear Case, Illustrated
             y                        y = g(x)
                                      y = f(x)

                     a   f(x) g(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
What then?

   But what if the func ons aren’t linear?
What then?

   But what if the func ons aren’t linear?
   Can we approximate a func on near a point with a linear
   func on?
What then?

   But what if the func ons aren’t linear?
   Can we approximate a func on near a point with a linear
   func on?
   What would be the slope of that linear func on?
What then?

   But what if the func ons aren’t linear?
   Can we approximate a func on near a point with a linear
   func on?
   What would be the slope of that linear func on? The
   deriva ve!
Theorem of the Day
 Theorem (L’Hopital’s Rule)
 Suppose f and g are differen able func ons 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 −∞.
Meet the Mathematician
  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 s pend to Johann
  Bernoulli, who proved this         Guillaume François Antoine,
  theorem and named it a er him!         Marquis de L’Hôpital
                                         (French, 1661–1704)
Revisiting the previous examples

 Example

               sin2 x
           lim
           x→0   x
Revisiting the previous examples

 Example

               sin2 x H     2 sin x cos x
           lim        = lim
           x→0   x      x→0       1
Revisiting the previous examples

 Example                              sin x → 0

               sin2 x H     2 sin x.cos x
           lim        = lim
           x→0   x      x→0       1
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
Revisiting the previous examples

 Example


        sin2 x
    lim
    x→0 sin x2
Revisiting the previous examples

 Example
             numerator → 0

        sin2 x.
    lim
    x→0 sin x2
Revisiting the previous examples

 Example
              numerator → 0

        sin2 x.
    lim        .
    x→0 sin x2




             denominator → 0
Revisiting the previous examples

 Example
             numerator → 0

        sin2 x. H       sin x cos x
                        2
    lim            lim
               . = x→0 (cos x2 ) (2x)
    x→0 sin x2                    




            denominator → 0
Revisiting the previous examples

 Example
                                 numerator → 0

        sin2 x H                  .
                      sin x cos x
                      2
    lim        = lim
    x→0 sin x2   x→0 (cos x2 ) (2x)
Revisiting the previous examples

 Example
                                  numerator → 0

        sin2 x H                   .
                      sin x cos x
                      2
    lim        = lim              .
    x→0 sin x2   x→0 (cos x2 ) (2x )
                                




                                denominator → 0
Revisiting the previous examples

 Example
                                 numerator → 0

        sin2 x H                   .        cos2 x − sin2 x
                      sin x cos x H
                      2
    lim        = lim                  lim
                                  . = x→0 cos x2 − 2x2 sin(x2 )
    x→0 sin x2   x→0 (cos x2 ) (2x )
                                




                               denominator → 0
Revisiting the previous examples

 Example
                                                      numerator → 1

        sin2 x H      sin x cos x H
                      2                     cos2 x − sin2 x.
    lim        = lim                = lim
    x→0 sin x2   x→0 (cos x2 ) (2x)
                                     x→0 cos x2 − 2x2 sin(x2 )
Revisiting the previous examples

 Example
                                                       numerator → 1

        sin2 x H      sin x cos x H
                      2                     cos2 x − sin2 x.
    lim        = lim                = lim                       .
    x→0 sin x2   x→0 (cos x2 ) (2x)
                                     x→0 cos x2 − 2x2 sin(x2 )




                                                      denominator → 1
Revisiting the previous examples

 Example


        sin2 x H      sin x cos x H
                      2                     cos2 x − sin2 x
    lim        = lim                = lim                       =1
    x→0 sin x2   x→0 (cos x2 ) (2x)
                                     x→0 cos x2 − 2x2 sin(x2 )
Revisiting the previous examples

 Example

              sin 3x
           lim
           x→0 sin x
Revisiting the previous examples

 Example

              sin 3x H     3 cos 3x
           lim       = lim          = 3.
           x→0 sin x   x→0 cos x
Another Example
 Example
 Find
                   x
             lim
             x→0 cos x
Beware of Red Herrings
 Example
 Find
                                      x
                                lim
                                x→0 cos x


 Solu on
 The limit of the denominator is 1, not 0, so L’Hôpital’s rule does not
 apply. The limit is 0.
Outline
 L’Hôpital’s Rule

 Rela ve Rates of Growth

 Other Indeterminate Limits
    Indeterminate Products
    Indeterminate Differences
    Indeterminate Powers
Limits of Rational Functions
revisited
 Example
          5x2 + 3x − 1
 Find lim 2            if it exists.
      x→∞ 3x + 7x + 27
Limits of Rational Functions
revisited
 Example
          5x2 + 3x − 1
 Find lim 2            if it exists.
      x→∞ 3x + 7x + 27


 Solu on
 Using L’Hôpital:

                5x2 + 3x − 1
            lim
           x→∞ 3x2 + 7x + 27
Limits of Rational Functions
revisited
 Example
          5x2 + 3x − 1
 Find lim 2            if it exists.
      x→∞ 3x + 7x + 27


 Solu on
 Using L’Hôpital:

                5x2 + 3x − 1 H     10x + 3
            lim 2            = lim
           x→∞ 3x + 7x + 27    x→∞ 6x + 7
Limits of Rational Functions
revisited
 Example
          5x2 + 3x − 1
 Find lim 2            if it exists.
      x→∞ 3x + 7x + 27


 Solu on
 Using L’Hôpital:

                5x2 + 3x − 1 H     10x + 3 H     10 5
            lim 2            = lim         = lim   =
           x→∞ 3x + 7x + 27    x→∞ 6x + 7    x→∞ 6   3
Limits of Rational Functions
revisited
 Example
          5x2 + 3x − 1
 Find lim 2            if it exists.
      x→∞ 3x + 7x + 27


 Solu on
 Using L’Hôpital:

                5x2 + 3x − 1 H     10x + 3 H     10 5
            lim 2            = lim         = lim   =
           x→∞ 3x + 7x + 27    x→∞ 6x + 7    x→∞ 6   3
Limits of Rational Functions
revisited II
 Example
          5x2 + 3x − 1
 Find lim              if it exists.
      x→∞   7x + 27
Limits of Rational Functions
revisited II
 Example
          5x2 + 3x − 1
 Find lim              if it exists.
      x→∞   7x + 27

 Solu on
 Using L’Hôpital:

                     5x2 + 3x − 1
                 lim
                x→∞    7x + 27
Limits of Rational Functions
revisited II
 Example
          5x2 + 3x − 1
 Find lim              if it exists.
      x→∞   7x + 27

 Solu on
 Using L’Hôpital:

                     5x2 + 3x − 1 H     10x + 3
                 lim              = lim
                x→∞    7x + 27      x→∞    7
Limits of Rational Functions
revisited II
 Example
          5x2 + 3x − 1
 Find lim              if it exists.
      x→∞   7x + 27

 Solu on
 Using L’Hôpital:

                     5x2 + 3x − 1 H     10x + 3
                 lim              = lim         =∞
                x→∞    7x + 27      x→∞    7
Limits of Rational Functions
revisited III
 Example
             4x + 7
 Find lim              if it exists.
     x→∞ 3x2 + 7x + 27
Limits of Rational Functions
revisited III
 Example
             4x + 7
 Find lim              if it exists.
     x→∞ 3x2 + 7x + 27


 Solu on
 Using L’Hôpital:
                        4x + 7
                lim
                x→∞ 3x2 + 7x + 27
Limits of Rational Functions
revisited III
 Example
             4x + 7
 Find lim              if it exists.
     x→∞ 3x2 + 7x + 27


 Solu on
 Using L’Hôpital:
                        4x + 7    H        4
                lim               = lim
                x→∞ 3x2 + 7x + 27   x→∞ 6x + 7
Limits of Rational Functions
revisited III
 Example
             4x + 7
 Find lim              if it exists.
     x→∞ 3x2 + 7x + 27


 Solu on
 Using L’Hôpital:
                        4x + 7    H        4
                lim               = lim        =0
                x→∞ 3x2 + 7x + 27   x→∞ 6x + 7
Limits of Rational Functions
 Fact
 Let f(x) and g(x) be polynomials of degree p and q.
                          f(x)
       If p  q, then lim      =∞
                      x→∞ g(x)
                          f(x)
       If p  q, then lim      =0
                      x→∞ g(x)
                          f(x)
       If p = q, then lim      is the ra o of the leading coefficients of
                      x→∞ g(x)
       f and g.
Exponential vs. geometric growth
 Example
         ex
 Find lim   , if it exists.
     x→∞ x2
Exponential vs. geometric growth
 Example
         ex
 Find lim   , if it exists.
     x→∞ x2


 Solu on
 We have
                        ex H     ex H     ex
                    lim    = lim    = lim    = ∞.
                   x→∞ x2    x→∞ 2x   x→∞ 2
Exponential vs. geometric growth
 Example
               ex
 What about lim   ?
           x→∞ x3
Exponential vs. geometric growth
 Example
                ex
 What about lim    ?
            x→∞ x3


 Answer
 S ll ∞. (Why?)
Exponential vs. geometric growth
 Example
                ex
 What about lim    ?
            x→∞ x3


 Answer
 S ll ∞. (Why?)

 Solu on

                ex H      ex H     ex H     ex
            lim    = lim 2 = lim      = lim    = ∞.
           x→∞ x3    x→∞ 3x    x→∞ 6x   x→∞ 6
Exponential vs. fractional powers
 Example
          ex
 Find lim √ , if it exists.
      x→∞  x
Exponential vs. fractional powers
 Example
          ex
 Find lim √ , if it exists.
      x→∞  x


 Solu on (without L’Hôpital)
 We have for all x  1, x1/2  x1 , so
                                 ex    ex
                                     
                                x1/2   x
 The right hand side tends to ∞, so the le -hand side must, too.
Exponential vs. fractional powers
 Example
          ex
 Find lim √ , if it exists.
      x→∞  x


 Solu on (with L’Hôpital)

                   ex        ex          √
               lim √ = lim 1 −1/2 = lim 2 xex = ∞
              x→∞   x x→∞ 2 x       x→∞
Exponential vs. any power
 Theorem
                                       ex
 Let r be any posi ve number. Then lim r = ∞.
                                   x→∞ x
Exponential vs. any power
 Theorem
                                       ex
 Let r be any posi ve number. Then lim r = ∞.
                                   x→∞ x


 Proof.
 If r is a posi ve integer, then apply L’Hôpital’s rule r mes to the frac-
   on. You get
                             ex H       H     ex
                        lim     = . . . = lim    = ∞.
                       x→∞ xr             x→∞ r!
Exponential vs. any power
 Theorem
                                       ex
 Let r be any posi ve number. Then lim r = ∞.
                                   x→∞ x


 Proof.
 If r is not an integer, let m be the smallest integer greater than r. Then
                          ex    ex
 if x  1, x  x , so r  m . The right-hand side tends to ∞ by the
              r    m
                          x     x
 previous step.
Any exponential vs. any power
 Theorem
                               ax
 Let a  1 and r  0. Then lim r = ∞.
                           x→∞ x
Any exponential vs. any power
 Theorem
                               ax
 Let a  1 and r  0. Then lim r = ∞.
                           x→∞ x

 Proof.
 If r is a posi ve integer, we have
                      ax H       H     (ln a)r ax
                  lim    = . . . = lim            = ∞.
                  x→∞ xr           x→∞     r!
 If r isn’t an integer, we can compare it as before.
Any exponential vs. any power
 Theorem
                               ax
 Let a  1 and r  0. Then lim r = ∞.
                           x→∞ x

 Proof.
 If r is a posi ve integer, we have
                      ax H       H     (ln a)r ax
                  lim    = . . . = lim            = ∞.
                  x→∞ xr           x→∞     r!
 If r isn’t an integer, we can compare it as before.
                 (1.00000001)x
 So even lim                     = ∞!
            x→∞     x100000000
Logarithmic versus power growth
 Theorem
                                     ln x
 Let r be any posi ve number. Then lim    = 0.
                                  x→∞ xr
Logarithmic versus power growth
 Theorem
                                      ln x
 Let r be any posi ve number. Then lim     = 0.
                                   x→∞ xr


 Proof.
 One applica on of L’Hôpital’s Rule here suffices:
                   ln x H      1/x       1
                lim     = lim r−1 = lim r = 0.
                x→∞ xr    x→∞ rx    x→∞ rx
Outline
 L’Hôpital’s Rule

 Rela ve Rates of Growth

 Other Indeterminate Limits
    Indeterminate Products
    Indeterminate Differences
    Indeterminate Powers
Indeterminate products
 Example
             √
 Find lim+    x ln x
      x→0

 This limit is of the form 0 · (−∞).
Indeterminate products
 Example
             √
 Find lim+    x ln x
      x→0

 This limit is of the form 0 · (−∞).
 Solu on
 Jury-rig the expression to make an indeterminate quo ent. Then
 apply L’Hôpital’s Rule:
             √
      lim        x ln x
     x→0+
Indeterminate products
 Example
             √
 Find lim+    x ln x
      x→0

 This limit is of the form 0 · (−∞).
 Solu on
 Jury-rig the expression to make an indeterminate quo ent. Then
 apply L’Hôpital’s Rule:
             √                 ln x
      lim        x ln x = lim+ 1 √
     x→0+                 x→0 / x
Indeterminate products
 Example
             √
 Find lim+    x ln x
      x→0

 This limit is of the form 0 · (−∞).
 Solu on
 Jury-rig the expression to make an indeterminate quo ent. Then
 apply L’Hôpital’s Rule:
             √                 ln x H       x−1
      lim        x ln x = lim+ 1 √ = lim+ 1 −3/2
     x→0+                 x→0 / x     x→0 − 2 x
Indeterminate products
 Example
             √
 Find lim+    x ln x
      x→0

 This limit is of the form 0 · (−∞).
 Solu on
 Jury-rig the expression to make an indeterminate quo ent. Then
 apply L’Hôpital’s Rule:
             √                 ln x H       x−1           √
      lim        x ln x = lim+ 1 √ = lim+ 1 −3/2 = lim+ −2 x
     x→0+                 x→0 / x     x→0 − 2 x    x→0
Indeterminate products
 Example
             √
 Find lim+    x ln x
      x→0

 This limit is of the form 0 · (−∞).
 Solu on
 Jury-rig the expression to make an indeterminate quo ent. Then
 apply L’Hôpital’s Rule:
             √                 ln x H       x−1           √
      lim        x ln x = lim+ 1 √ = lim+ 1 −3/2 = lim+ −2 x = 0
     x→0+                 x→0 / x     x→0 − 2 x    x→0
Indeterminate differences

 Example
                                (                )
                                    1
                          lim         − cot 2x
                         x→0+       x

 This limit is of the form ∞ − ∞.
Indeterminate Differences
 Solu on
 Again, rig it to make an indeterminate quo ent.
                  (           )
                    1                  sin(2x) − x cos(2x)
             lim+     − cot 2x = lim+
            x→0     x             x→0       x sin(2x)
Indeterminate Differences
 Solu on
 Again, rig it to make an indeterminate quo ent.
                  (           )
                    1                  sin(2x) − x cos(2x)
             lim+     − cot 2x = lim+
            x→0     x             x→0       x sin(2x)
                                H      cos(2x) + 2x sin(2x)
                                = lim+
                                  x→0 2x cos(2x) + sin(2x)
Indeterminate Differences
 Solu on
 Again, rig it to make an indeterminate quo ent.
                  (           )
                    1                  sin(2x) − x cos(2x)
             lim+     − cot 2x = lim+
            x→0     x             x→0       x sin(2x)
                                H      cos(2x) + 2x sin(2x)
                                = lim+
                                  x→0 2x cos(2x) + sin(2x)
                                =∞
Indeterminate Differences
 Solu on
 Again, rig it to make an indeterminate quo ent.
                  (           )
                    1                  sin(2x) − x cos(2x)
             lim+     − cot 2x = lim+
            x→0     x             x→0       x sin(2x)
                                H      cos(2x) + 2x sin(2x)
                                = lim+
                                  x→0 2x cos(2x) + sin(2x)
                                =∞

 The limit is +∞ because the numerator tends to 1 while the
 denominator tends to zero but remains posi ve.
Checking your work
This all goes in the thought cloud

    tan 2x                                               1
 lim       = 1, so for small x, tan 2x ≈ 2x. So cot 2x ≈    and
 x→0 2x                                                  2x
                  1           1  1   1
                    − cot 2x ≈ −   =   →∞
                  x           x 2x 2x
 as x → 0+ .
Indeterminate powers
 Example
 Find lim+ (1 − 2x)1/x
      x→0
Indeterminate powers
 Example
 Find lim+ (1 − 2x)1/x
      x→0


 Solu on
 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
Indeterminate powers
 Example
 Find lim+ (1 − 2x)1/x
      x→0


 Solu on
                          0
 This limit is of the form , so we can use L’Hôpital:
                          0
                                            −2
                        ln(1 − 2x) H
                   lim+            = lim+ 1−2x = −2
                  x→0        x       x→0   1
Indeterminate powers
 Example
 Find lim+ (1 − 2x)1/x
      x→0


 Solu on
                          0
 This limit is of the form , so we can use L’Hôpital:
                          0
                                            −2
                        ln(1 − 2x) H
                   lim+            = lim+ 1−2x = −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 .
Another indeterminate power limit
 Example
 Find lim+ (3x)4x
      x→0
Another indeterminate power limit
 Example
 Find lim+ (3x)4x
      x→0


 Solu on

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


 So the answer is e0 = 1.
Summary
  Form   Method
    0
    0    L’Hôpital’s rule directly
   ∞
   ∞     L’Hôpital’s rule directly
                               ∞
  0·∞    jiggle to make 0 or
                        0      ∞.

 ∞ − ∞ combine to make an indeterminate product or quo ent
   00    take ln to make an indeterminate product
  ∞0     di o
   1∞    di o
Final Thoughts

   L’Hôpital’s Rule only works on indeterminate quo ents
   Luckily, most indeterminate limits can be transformed into
   indeterminate quo ents
   L’Hôpital’s Rule gives wrong answers for non-indeterminate
   limits!

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Lesson 17: Indeterminate forms and l'Hôpital's Rule (slides)

  • 1. Sec on 3.7 Indeterminate forms and lHôpital’s Rule V63.0121.011: Calculus I Professor Ma hew Leingang New York University March 30, 2011 .
  • 2. Announcements Midterm has been returned. Please see FAQ on Blackboard (under ”Exams and Quizzes”) Quiz 3 this week in recita on on Sec on 2.6, 2.8, 3.1, 3.2
  • 3. Objectives Know when a limit is of indeterminate form: indeterminate quo ents: 0/0, ∞/∞ indeterminate products: 0×∞ indeterminate differences: ∞ − ∞ indeterminate powers: 00 , ∞0 , and 1∞ Resolve limits in indeterminate form
  • 4. Recall Recall the limit laws from Chapter 2. Limit of a sum is the sum of the limits
  • 5. 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
  • 6. 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
  • 7. 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 quo ent is the quo ent of the limits ... whoops! This is true as long as you don’t try to divide by zero.
  • 8. More about dividing limits We know dividing by zero is bad. Most of the me, if an expression’s numerator approaches a finite nonzero number and denominator approaches zero, the quo ent has an infinite. For example: 1 cos x lim+ = +∞ lim− = −∞ x→0 x x→0 x3
  • 9. Why 1/0 ̸= ∞ 1 Consider the func on f(x) = 1 . x sin x y . x Then lim f(x) is of the form 1/0, but the limit does not exist and is x→∞ not infinite.
  • 10. Why 1/0 ̸= ∞ 1 Consider the func on f(x) = 1 . x sin x y . x Then lim f(x) is of the form 1/0, but the limit does not exist and is x→∞ not infinite. Even less predictable: when numerator and denominator both go to zero.
  • 11. Experiments with funny limits sin2 x lim x→0 x .
  • 12. Experiments with funny limits sin2 x lim =0 x→0 x .
  • 13. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 x→0 sin x .
  • 14. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 does not exist x→0 sin x .
  • 15. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 does not exist x→0 sin x sin2 x . lim x→0 sin(x2 )
  • 16. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 does not exist x→0 sin x sin2 x . lim =1 x→0 sin(x2 )
  • 17. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 does not exist x→0 sin x sin2 x . lim =1 x→0 sin(x2 ) sin 3x lim x→0 sin x
  • 18. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 does not exist x→0 sin x sin2 x . lim =1 x→0 sin(x2 ) sin 3x lim =3 x→0 sin x
  • 19. Experiments with funny limits sin2 x lim =0 x→0 x x lim 2 does not exist x→0 sin x sin2 x . lim =1 x→0 sin(x2 ) sin 3x lim =3 x→0 sin x 0 All of these are of the form , and since we can get different 0 answers in different cases, we say this form is indeterminate.
  • 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 0 case “is” , and therefore nonexistent 0 because this expression is undefined. 0 The limit is of the form , which means 0 we cannot evaluate it with our limit laws.
  • 21. Indeterminate forms are like Tug Of War Which side wins depends on which side is stronger.
  • 22. Outline L’Hôpital’s Rule Rela ve Rates of Growth Other Indeterminate Limits Indeterminate Products Indeterminate Differences Indeterminate Powers
  • 23. The Linear Case Ques on f(x) If f and g are lines and f(a) = g(a) = 0, what is lim ? x→a g(x)
  • 24. The Linear Case Ques on f(x) If f and g are lines and f(a) = g(a) = 0, what is lim ? x→a g(x) Solu on The func ons f and g can be wri en in the form f(x) = m1 (x − a) g(x) = m2 (x − a) So f(x) m1 = g(x) m2
  • 25. The Linear Case, Illustrated y y = g(x) y = f(x) a f(x) g(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
  • 26. What then? But what if the func ons aren’t linear?
  • 27. What then? But what if the func ons aren’t linear? Can we approximate a func on near a point with a linear func on?
  • 28. What then? But what if the func ons aren’t linear? Can we approximate a func on near a point with a linear func on? What would be the slope of that linear func on?
  • 29. What then? But what if the func ons aren’t linear? Can we approximate a func on near a point with a linear func on? What would be the slope of that linear func on? The deriva ve!
  • 30. Theorem of the Day Theorem (L’Hopital’s Rule) Suppose f and g are differen able func ons 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 −∞.
  • 31. Meet the Mathematician 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 s pend to Johann Bernoulli, who proved this Guillaume François Antoine, theorem and named it a er him! Marquis de L’Hôpital (French, 1661–1704)
  • 32. Revisiting the previous examples Example sin2 x lim x→0 x
  • 33. Revisiting the previous examples Example sin2 x H 2 sin x cos x lim = lim x→0 x x→0 1
  • 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
  • 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
  • 36. Revisiting the previous examples Example sin2 x lim x→0 sin x2
  • 37. Revisiting the previous examples Example numerator → 0 sin2 x. lim x→0 sin x2
  • 38. Revisiting the previous examples Example numerator → 0 sin2 x. lim . x→0 sin x2 denominator → 0
  • 39. Revisiting the previous examples Example numerator → 0 sin2 x. H sin x cos x 2 lim lim . = x→0 (cos x2 ) (2x) x→0 sin x2 denominator → 0
  • 40. Revisiting the previous examples Example numerator → 0 sin2 x H . sin x cos x 2 lim = lim x→0 sin x2 x→0 (cos x2 ) (2x)
  • 41. Revisiting the previous examples Example numerator → 0 sin2 x H . sin x cos x 2 lim = lim . x→0 sin x2 x→0 (cos x2 ) (2x ) denominator → 0
  • 42. Revisiting the previous examples Example numerator → 0 sin2 x H . cos2 x − sin2 x sin x cos x H 2 lim = lim lim . = x→0 cos x2 − 2x2 sin(x2 ) x→0 sin x2 x→0 (cos x2 ) (2x ) denominator → 0
  • 43. Revisiting the previous examples Example numerator → 1 sin2 x H sin x cos x H 2 cos2 x − sin2 x. lim = lim = lim x→0 sin x2 x→0 (cos x2 ) (2x) x→0 cos x2 − 2x2 sin(x2 )
  • 44. Revisiting the previous examples Example numerator → 1 sin2 x H sin x cos x H 2 cos2 x − sin2 x. lim = lim = lim . x→0 sin x2 x→0 (cos x2 ) (2x) x→0 cos x2 − 2x2 sin(x2 ) denominator → 1
  • 45. Revisiting the previous examples Example sin2 x H sin x cos x H 2 cos2 x − sin2 x lim = lim = lim =1 x→0 sin x2 x→0 (cos x2 ) (2x) x→0 cos x2 − 2x2 sin(x2 )
  • 46. Revisiting the previous examples Example sin 3x lim x→0 sin x
  • 47. Revisiting the previous examples Example sin 3x H 3 cos 3x lim = lim = 3. x→0 sin x x→0 cos x
  • 48. Another Example Example Find x lim x→0 cos x
  • 49. Beware of Red Herrings Example Find x lim x→0 cos x Solu on The limit of the denominator is 1, not 0, so L’Hôpital’s rule does not apply. The limit is 0.
  • 50. Outline L’Hôpital’s Rule Rela ve Rates of Growth Other Indeterminate Limits Indeterminate Products Indeterminate Differences Indeterminate Powers
  • 51. Limits of Rational Functions revisited Example 5x2 + 3x − 1 Find lim 2 if it exists. x→∞ 3x + 7x + 27
  • 52. Limits of Rational Functions revisited Example 5x2 + 3x − 1 Find lim 2 if it exists. x→∞ 3x + 7x + 27 Solu on Using L’Hôpital: 5x2 + 3x − 1 lim x→∞ 3x2 + 7x + 27
  • 53. Limits of Rational Functions revisited Example 5x2 + 3x − 1 Find lim 2 if it exists. x→∞ 3x + 7x + 27 Solu on Using L’Hôpital: 5x2 + 3x − 1 H 10x + 3 lim 2 = lim x→∞ 3x + 7x + 27 x→∞ 6x + 7
  • 54. Limits of Rational Functions revisited Example 5x2 + 3x − 1 Find lim 2 if it exists. x→∞ 3x + 7x + 27 Solu on Using L’Hôpital: 5x2 + 3x − 1 H 10x + 3 H 10 5 lim 2 = lim = lim = x→∞ 3x + 7x + 27 x→∞ 6x + 7 x→∞ 6 3
  • 55. Limits of Rational Functions revisited Example 5x2 + 3x − 1 Find lim 2 if it exists. x→∞ 3x + 7x + 27 Solu on Using L’Hôpital: 5x2 + 3x − 1 H 10x + 3 H 10 5 lim 2 = lim = lim = x→∞ 3x + 7x + 27 x→∞ 6x + 7 x→∞ 6 3
  • 56. Limits of Rational Functions revisited II Example 5x2 + 3x − 1 Find lim if it exists. x→∞ 7x + 27
  • 57. Limits of Rational Functions revisited II Example 5x2 + 3x − 1 Find lim if it exists. x→∞ 7x + 27 Solu on Using L’Hôpital: 5x2 + 3x − 1 lim x→∞ 7x + 27
  • 58. Limits of Rational Functions revisited II Example 5x2 + 3x − 1 Find lim if it exists. x→∞ 7x + 27 Solu on Using L’Hôpital: 5x2 + 3x − 1 H 10x + 3 lim = lim x→∞ 7x + 27 x→∞ 7
  • 59. Limits of Rational Functions revisited II Example 5x2 + 3x − 1 Find lim if it exists. x→∞ 7x + 27 Solu on Using L’Hôpital: 5x2 + 3x − 1 H 10x + 3 lim = lim =∞ x→∞ 7x + 27 x→∞ 7
  • 60. Limits of Rational Functions revisited III Example 4x + 7 Find lim if it exists. x→∞ 3x2 + 7x + 27
  • 61. Limits of Rational Functions revisited III Example 4x + 7 Find lim if it exists. x→∞ 3x2 + 7x + 27 Solu on Using L’Hôpital: 4x + 7 lim x→∞ 3x2 + 7x + 27
  • 62. Limits of Rational Functions revisited III Example 4x + 7 Find lim if it exists. x→∞ 3x2 + 7x + 27 Solu on Using L’Hôpital: 4x + 7 H 4 lim = lim x→∞ 3x2 + 7x + 27 x→∞ 6x + 7
  • 63. Limits of Rational Functions revisited III Example 4x + 7 Find lim if it exists. x→∞ 3x2 + 7x + 27 Solu on Using L’Hôpital: 4x + 7 H 4 lim = lim =0 x→∞ 3x2 + 7x + 27 x→∞ 6x + 7
  • 64. Limits of Rational Functions Fact Let f(x) and g(x) be polynomials of degree p and q. f(x) If p q, then lim =∞ x→∞ g(x) f(x) If p q, then lim =0 x→∞ g(x) f(x) If p = q, then lim is the ra o of the leading coefficients of x→∞ g(x) f and g.
  • 65. Exponential vs. geometric growth Example ex Find lim , if it exists. x→∞ x2
  • 66. Exponential vs. geometric growth Example ex Find lim , if it exists. x→∞ x2 Solu on We have ex H ex H ex lim = lim = lim = ∞. x→∞ x2 x→∞ 2x x→∞ 2
  • 67. Exponential vs. geometric growth Example ex What about lim ? x→∞ x3
  • 68. Exponential vs. geometric growth Example ex What about lim ? x→∞ x3 Answer S ll ∞. (Why?)
  • 69. Exponential vs. geometric growth Example ex What about lim ? x→∞ x3 Answer S ll ∞. (Why?) Solu on ex H ex H ex H ex lim = lim 2 = lim = lim = ∞. x→∞ x3 x→∞ 3x x→∞ 6x x→∞ 6
  • 70. Exponential vs. fractional powers Example ex Find lim √ , if it exists. x→∞ x
  • 71. Exponential vs. fractional powers Example ex Find lim √ , if it exists. x→∞ x Solu on (without L’Hôpital) We have for all x 1, x1/2 x1 , so ex ex x1/2 x The right hand side tends to ∞, so the le -hand side must, too.
  • 72. Exponential vs. fractional powers Example ex Find lim √ , if it exists. x→∞ x Solu on (with L’Hôpital) ex ex √ lim √ = lim 1 −1/2 = lim 2 xex = ∞ x→∞ x x→∞ 2 x x→∞
  • 73. Exponential vs. any power Theorem ex Let r be any posi ve number. Then lim r = ∞. x→∞ x
  • 74. Exponential vs. any power Theorem ex Let r be any posi ve number. Then lim r = ∞. x→∞ x Proof. If r is a posi ve integer, then apply L’Hôpital’s rule r mes to the frac- on. You get ex H H ex lim = . . . = lim = ∞. x→∞ xr x→∞ r!
  • 75. Exponential vs. any power Theorem ex Let r be any posi ve number. Then lim r = ∞. x→∞ x Proof. If r is not an integer, let m be the smallest integer greater than r. Then ex ex if x 1, x x , so r m . The right-hand side tends to ∞ by the r m x x previous step.
  • 76. Any exponential vs. any power Theorem ax Let a 1 and r 0. Then lim r = ∞. x→∞ x
  • 77. Any exponential vs. any power Theorem ax Let a 1 and r 0. Then lim r = ∞. x→∞ x Proof. If r is a posi ve integer, we have ax H H (ln a)r ax lim = . . . = lim = ∞. x→∞ xr x→∞ r! If r isn’t an integer, we can compare it as before.
  • 78. Any exponential vs. any power Theorem ax Let a 1 and r 0. Then lim r = ∞. x→∞ x Proof. If r is a posi ve integer, we have ax H H (ln a)r ax lim = . . . = lim = ∞. x→∞ xr x→∞ r! If r isn’t an integer, we can compare it as before. (1.00000001)x So even lim = ∞! x→∞ x100000000
  • 79. Logarithmic versus power growth Theorem ln x Let r be any posi ve number. Then lim = 0. x→∞ xr
  • 80. Logarithmic versus power growth Theorem ln x Let r be any posi ve number. Then lim = 0. x→∞ xr Proof. One applica on of L’Hôpital’s Rule here suffices: ln x H 1/x 1 lim = lim r−1 = lim r = 0. x→∞ xr x→∞ rx x→∞ rx
  • 81. Outline L’Hôpital’s Rule Rela ve Rates of Growth Other Indeterminate Limits Indeterminate Products Indeterminate Differences Indeterminate Powers
  • 82. Indeterminate products Example √ Find lim+ x ln x x→0 This limit is of the form 0 · (−∞).
  • 83. Indeterminate products Example √ Find lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solu on Jury-rig the expression to make an indeterminate quo ent. Then apply L’Hôpital’s Rule: √ lim x ln x x→0+
  • 84. Indeterminate products Example √ Find lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solu on Jury-rig the expression to make an indeterminate quo ent. Then apply L’Hôpital’s Rule: √ ln x lim x ln x = lim+ 1 √ x→0+ x→0 / x
  • 85. Indeterminate products Example √ Find lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solu on Jury-rig the expression to make an indeterminate quo ent. Then apply L’Hôpital’s Rule: √ ln x H x−1 lim x ln x = lim+ 1 √ = lim+ 1 −3/2 x→0+ x→0 / x x→0 − 2 x
  • 86. Indeterminate products Example √ Find lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solu on Jury-rig the expression to make an indeterminate quo ent. Then apply L’Hôpital’s Rule: √ ln x H x−1 √ lim x ln x = lim+ 1 √ = lim+ 1 −3/2 = lim+ −2 x x→0+ x→0 / x x→0 − 2 x x→0
  • 87. Indeterminate products Example √ Find lim+ x ln x x→0 This limit is of the form 0 · (−∞). Solu on Jury-rig the expression to make an indeterminate quo ent. Then apply L’Hôpital’s Rule: √ ln x H x−1 √ lim x ln x = lim+ 1 √ = lim+ 1 −3/2 = lim+ −2 x = 0 x→0+ x→0 / x x→0 − 2 x x→0
  • 88. Indeterminate differences Example ( ) 1 lim − cot 2x x→0+ x This limit is of the form ∞ − ∞.
  • 89. Indeterminate Differences Solu on Again, rig it to make an indeterminate quo ent. ( ) 1 sin(2x) − x cos(2x) lim+ − cot 2x = lim+ x→0 x x→0 x sin(2x)
  • 90. Indeterminate Differences Solu on Again, rig it to make an indeterminate quo ent. ( ) 1 sin(2x) − x cos(2x) lim+ − cot 2x = lim+ x→0 x x→0 x sin(2x) H cos(2x) + 2x sin(2x) = lim+ x→0 2x cos(2x) + sin(2x)
  • 91. Indeterminate Differences Solu on Again, rig it to make an indeterminate quo ent. ( ) 1 sin(2x) − x cos(2x) lim+ − cot 2x = lim+ x→0 x x→0 x sin(2x) H cos(2x) + 2x sin(2x) = lim+ x→0 2x cos(2x) + sin(2x) =∞
  • 92. Indeterminate Differences Solu on Again, rig it to make an indeterminate quo ent. ( ) 1 sin(2x) − x cos(2x) lim+ − cot 2x = lim+ x→0 x x→0 x sin(2x) H cos(2x) + 2x sin(2x) = lim+ x→0 2x cos(2x) + sin(2x) =∞ The limit is +∞ because the numerator tends to 1 while the denominator tends to zero but remains posi ve.
  • 93. Checking your work This all goes in the thought cloud tan 2x 1 lim = 1, so for small x, tan 2x ≈ 2x. So cot 2x ≈ and x→0 2x 2x 1 1 1 1 − cot 2x ≈ − = →∞ x x 2x 2x as x → 0+ .
  • 94. Indeterminate powers Example Find lim+ (1 − 2x)1/x x→0
  • 95. Indeterminate powers Example Find lim+ (1 − 2x)1/x x→0 Solu on 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
  • 96. Indeterminate powers Example Find lim+ (1 − 2x)1/x x→0 Solu on 0 This limit is of the form , so we can use L’Hôpital: 0 −2 ln(1 − 2x) H lim+ = lim+ 1−2x = −2 x→0 x x→0 1
  • 97. Indeterminate powers Example Find lim+ (1 − 2x)1/x x→0 Solu on 0 This limit is of the form , so we can use L’Hôpital: 0 −2 ln(1 − 2x) H lim+ = lim+ 1−2x = −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 .
  • 98. Another indeterminate power limit Example Find lim+ (3x)4x x→0
  • 99. Another indeterminate power limit Example Find lim+ (3x)4x x→0 Solu on ln(3x) ln lim+ (3x)4x = lim+ ln(3x)4x = lim+ 4x ln(3x) = lim+ 1/4x x→0 x→0 x→0 x→0 H 3/3x = lim+ −1 2 = lim+ (−4x) = 0 x→0 /4x x→0 So the answer is e0 = 1.
  • 100. Summary Form Method 0 0 L’Hôpital’s rule directly ∞ ∞ L’Hôpital’s rule directly ∞ 0·∞ jiggle to make 0 or 0 ∞. ∞ − ∞ combine to make an indeterminate product or quo ent 00 take ln to make an indeterminate product ∞0 di o 1∞ di o
  • 101. Final Thoughts L’Hôpital’s Rule only works on indeterminate quo ents Luckily, most indeterminate limits can be transformed into indeterminate quo ents L’Hôpital’s Rule gives wrong answers for non-indeterminate limits!