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Section 2.2–3
                  The Concept of Limit
                      Limit Laws

                             Math 1a


                         February 4, 2008


Announcements
   Syllabus available on course website
   Homework for Wednesday 2/6:
       Practice 2.2: 1, 3, 5, 7, 13, 15; 2.3: 1, 3, 7, 13, 15, 17
       Turn-in 2:2: 2, 4, 6, 8; 2.3: 2, 20, 38
   Homework for Monday 2/11: 2.2.28, 2.3.30, 2.4.34
   ALEKS due Wednesday 2/20
Outline


   The Concept of Limit
      Heuristics
      Errors and tolerances
      Pathologies


   Limit Laws
      Easy laws
      The direct substitution property
      Limits by algebra
      Two more limit theorems
Zeno’s Paradox




                     That which is in
                     locomotion must
                     arrive at the
                     half-way stage
                     before it arrives at
                     the goal.
                 (Aristotle Physics VI:9,
                 239b10)
Heuristic Definition of a Limit



   Definition
   We write
                                lim f (x) = L
                               x→a

   and say

              “the limit of f (x), as x approaches a, equals L”

   if we can make the values of f (x) arbitrarily close to L (as close to
   L as we like) by taking x to be sufficiently close to a (on either side
   of a) but not equal to a.
The error-tolerance game




     L




                           a
The error-tolerance game




     L




                           a
The error-tolerance game




     L




                           a
The error-tolerance game



                     This tolerance is too big

     L




                           a
The error-tolerance game




     L




                           a
The error-tolerance game



                           Still too big

     L




                           a
The error-tolerance game




     L




                           a
The error-tolerance game



                           This looks good

     L




                            a
The error-tolerance game



                           So does this

     L




                           a
Examples


  Example
  Find lim x 2 if it exists.
       x→0
Examples


  Example
  Find lim x 2 if it exists.
       x→0

  Example
          |x|
  Find lim    if it exists.
       x→0 x
Examples


  Example
  Find lim x 2 if it exists.
       x→0

  Example
          |x|
  Find lim    if it exists.
       x→0 x

  Example
              1
  Find lim+     if it exists.
       x→0    x
Examples


  Example
  Find lim x 2 if it exists.
       x→0

  Example
          |x|
  Find lim    if it exists.
       x→0 x

  Example
              1
  Find lim+     if it exists.
       x→0    x
  Example
                  π
  Find lim sin        if it exists.
       x→0        x
What could go wrong?




  How could a function fail to have a limit? Some possibilities:
      left- and right- hand limits exist but are not equal
      The function is unbounded near a
      Oscillation with increasingly high frequency near a
Precise Definition of a Limit



   Let f be a function defined on an some open interval that contains
   the number a, except possibly at a itself. Then we say that the
   limit of f (x) as x approaches a is L, and we write

                              lim f (x) = L,
                              x→a

   if for every ε > 0 there is a corresponding δ > 0 such that

                 if 0 < |x − a| < δ, then |f (x) − L| < ε.
The error-tolerance game = ε, δ




     L




                          a
The error-tolerance game = ε, δ




   L+ε
    L
   L−ε




                          a
The error-tolerance game = ε, δ




   L+ε
    L
   L−ε




                      a − δaa + δ
The error-tolerance game = ε, δ



                         This δ is too big
   L+ε
    L
   L−ε




                      a − δaa + δ
The error-tolerance game = ε, δ




   L+ε
    L
   L−ε




                        a −aδ δ
                          a+
The error-tolerance game = ε, δ



                         This δ looks good
   L+ε
    L
   L−ε




                        a −aδ δ
                          a+
The error-tolerance game = ε, δ



                          So does this δ
   L+ε
    L
   L−ε




                        aa a δ δ
                         − +
Meet the Mathematician: Augustin Louis Cauchy



     French, 1789–1857
     Royalist and Catholic
     made contributions in
     geometry, calculus,
     complex analysis,
     number theory
     created the definition of
     limit we use today but
     didn’t understand it
Outline


   The Concept of Limit
      Heuristics
      Errors and tolerances
      Pathologies


   Limit Laws
      Easy laws
      The direct substitution property
      Limits by algebra
      Two more limit theorems
Limit Laws


   Suppose that c is a constant and the limits

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

   exist. Then
   1. lim [f (x) + g (x)] = lim f (x) + lim g (x)
      x→a                      x→a         x→a
   2. lim [f (x) − g (x)] = lim f (x) − lim g (x)
      x→a                      x→a         x→a
   3. lim [cf (x)] = c lim f (x)
      x→a               x→a
   4. lim [f (x)g (x)] = lim f (x) · lim g (x)
      x→a                    x→a     x→a
Limit Laws, continued

         f (x)   lim f (x)
   5. lim      = x→a       , if lim g (x) = 0.
     x→a g (x)   lim g (x)      x→a
                  x→a
Limit Laws, continued

          f (x)   lim f (x)
   5. lim       = x→a       , if lim g (x) = 0.
      x→a g (x)   lim g (x)      x→a
                    x→a
                                n
                n
   6. lim [f (x)] = lim f (x)
      x→a             x→a
Limit Laws, continued

          f (x)   lim f (x)
   5. lim       = x→a       , if lim g (x) = 0.
      x→a g (x)   lim g (x)      x→a
                    x→a
                                n
                n
   6. lim [f (x)] = lim f (x)       (follows from 3 repeatedly)
      x→a             x→a
Limit Laws, continued

          f (x)   lim f (x)
   5. lim       = x→a       , if lim g (x) = 0.
      x→a g (x)   lim g (x)      x→a
                      x→a
                                n
                  n
   6. lim [f (x)] = lim f (x)       (follows from 3 repeatedly)
      x→a               x→a
   7. lim c = c
      x→a
   8. lim x = a
      x→a
Limit Laws, continued

          f (x)   lim f (x)
   5. lim       = x→a       , if lim g (x) = 0.
      x→a g (x)   lim g (x)      x→a
                      x→a
                                n
                  n
   6. lim [f (x)] = lim f (x)       (follows from 3 repeatedly)
      x→a               x→a
   7. lim c = c
      x→a
   8. lim x = a
      x→a
   9. lim x n = an
      x→a
          √      √
  10. lim n x = n a
      x→a
Limit Laws, continued

          f (x)   lim f (x)
   5. lim       = x→a       , if lim g (x) = 0.
      x→a g (x)   lim g (x)      x→a
                      x→a
                                n
                  n
   6. lim [f (x)] = lim f (x)       (follows from 3 repeatedly)
      x→a               x→a
   7. lim c = c
      x→a
   8. lim x = a
      x→a
   9. lim x n = an (follows from 6 and 8)
      x→a
          √      √
  10. lim n x = n a
      x→a
Limit Laws, continued

          f (x)   lim f (x)
   5. lim       = x→a       , if lim g (x) = 0.
      x→a g (x)   lim g (x)      x→a
                       x→a
                                      n
                   n
   6. lim [f (x)] = lim f (x)             (follows from 3 repeatedly)
      x→a                 x→a
   7. lim c = c
      x→a
   8. lim x = a
      x→a
   9. lim x n = an (follows from 6 and 8)
      x→a
          √      √
  10. lim n x = n a
      x→a
            n
  11. lim       f (x) =   n   lim f (x) (If n is even, we must additionally
      x→a                     x→a
      assume that lim f (x) > 0)
                       x→a
Direct Substitution Property




   Theorem (The Direct Substitution Property)
   If f is a polynomial or a rational function and a is in the domain of
   f , then
                              lim f (x) = f (a)
                             x→a
Limits do not see the point! (in a good way)


   Theorem
   If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L.
                                    x→a                 x→a
Limits do not see the point! (in a good way)


   Theorem
   If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L.
                                      x→a               x→a

   Example
             x 2 + 2x + 1
   Find lim               , if it exists.
        x→−1     x +1
Limits do not see the point! (in a good way)


   Theorem
   If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L.
                                      x→a               x→a

   Example
             x 2 + 2x + 1
   Find lim               , if it exists.
        x→−1     x +1
   Solution
         x 2 + 2x + 1
   Since              = x + 1 whenever x = −1, and since
             x +1
                                 x 2 + 2x + 1
    lim x + 1 = 0, we have lim                = 0.
   x→−1                     x→−1     x +1
Finding limits by algebraic manipulations

   Example √
               x −2
   Find lim         .
       x→4    x −4
Finding limits by algebraic manipulations

   Example √
               x −2
   Find lim         .
       x→4    x −4
   Solution                           √            √       √
                                           2
   Write the denominator as x − 4 =       x − 4 = ( x − 2)( x + 2).
Finding limits by algebraic manipulations

   Example √
               x −2
   Find lim         .
       x→4    x −4
   Solution                         √ 2        √     √
   Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2).
   So
                     √                  √
                       x −2               x −2
                 lim         = lim √         √
                x→2 x − 4      x→2 ( x − 2)( x + 2)
                                      1      1
                             = lim √       =
                               x→2   x +2    4
Finding limits by algebraic manipulations

   Example √
                   x −2
   Find lim             .
       x→4        x −4
   Solution                         √ 2        √     √
   Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2).
   So
                     √                  √
                       x −2               x −2
                 lim         = lim √         √
                x→2 x − 4      x→2 ( x − 2)( x + 2)
                                      1      1
                             = lim √       =
                               x→2   x +2    4


   Example√          √
              3
                  x− 3a
   Try lim              .
      x→a         x −a
Two More Important Limit Theorems

  Theorem
  If f (x) ≤ g (x) when x is near a (except possibly at a), then

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

  (as usual, provided these limits exist).

  Theorem (The Squeeze/Sandwich/Pinching Theorem)
  If f (x) ≤ g (x) ≤ h(x) when x is near a (as usual, except possibly
  at a), and
                        lim f (x) = lim h(x) = L,
                       x→a          x→a

  then
                              lim g (x) = L.
                              x→a
We can use the Squeeze Theorem to make complicated limits
simple.
We can use the Squeeze Theorem to make complicated limits
simple.
Example
                        1
Show that lim x 2 sin       = 0.
           x→0          x
We can use the Squeeze Theorem to make complicated limits
simple.
Example
                         1
Show that lim x 2 sin         = 0.
           x→0           x
Solution
We have for all x,
                                         1
                        −x 2 ≤ x 2 sin       ≤ x2
                                         x

The left and right sides go to zero as x → 0.

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Lesson 2: Limits and Limit Laws

  • 1. Section 2.2–3 The Concept of Limit Limit Laws Math 1a February 4, 2008 Announcements Syllabus available on course website Homework for Wednesday 2/6: Practice 2.2: 1, 3, 5, 7, 13, 15; 2.3: 1, 3, 7, 13, 15, 17 Turn-in 2:2: 2, 4, 6, 8; 2.3: 2, 20, 38 Homework for Monday 2/11: 2.2.28, 2.3.30, 2.4.34 ALEKS due Wednesday 2/20
  • 2. Outline The Concept of Limit Heuristics Errors and tolerances Pathologies Limit Laws Easy laws The direct substitution property Limits by algebra Two more limit theorems
  • 3. Zeno’s Paradox That which is in locomotion must arrive at the half-way stage before it arrives at the goal. (Aristotle Physics VI:9, 239b10)
  • 4. Heuristic Definition of a Limit Definition We write lim f (x) = L x→a and say “the limit of f (x), as x approaches a, equals L” if we can make the values of f (x) arbitrarily close to L (as close to L as we like) by taking x to be sufficiently close to a (on either side of a) but not equal to a.
  • 8. The error-tolerance game This tolerance is too big L a
  • 10. The error-tolerance game Still too big L a
  • 12. The error-tolerance game This looks good L a
  • 13. The error-tolerance game So does this L a
  • 14. Examples Example Find lim x 2 if it exists. x→0
  • 15. Examples Example Find lim x 2 if it exists. x→0 Example |x| Find lim if it exists. x→0 x
  • 16. Examples Example Find lim x 2 if it exists. x→0 Example |x| Find lim if it exists. x→0 x Example 1 Find lim+ if it exists. x→0 x
  • 17. Examples Example Find lim x 2 if it exists. x→0 Example |x| Find lim if it exists. x→0 x Example 1 Find lim+ if it exists. x→0 x Example π Find lim sin if it exists. x→0 x
  • 18. What could go wrong? How could a function fail to have a limit? Some possibilities: left- and right- hand limits exist but are not equal The function is unbounded near a Oscillation with increasingly high frequency near a
  • 19. Precise Definition of a Limit Let f be a function defined on an some open interval that contains the number a, except possibly at a itself. Then we say that the limit of f (x) as x approaches a is L, and we write lim f (x) = L, x→a if for every ε > 0 there is a corresponding δ > 0 such that if 0 < |x − a| < δ, then |f (x) − L| < ε.
  • 21. The error-tolerance game = ε, δ L+ε L L−ε a
  • 22. The error-tolerance game = ε, δ L+ε L L−ε a − δaa + δ
  • 23. The error-tolerance game = ε, δ This δ is too big L+ε L L−ε a − δaa + δ
  • 24. The error-tolerance game = ε, δ L+ε L L−ε a −aδ δ a+
  • 25. The error-tolerance game = ε, δ This δ looks good L+ε L L−ε a −aδ δ a+
  • 26. The error-tolerance game = ε, δ So does this δ L+ε L L−ε aa a δ δ − +
  • 27. Meet the Mathematician: Augustin Louis Cauchy French, 1789–1857 Royalist and Catholic made contributions in geometry, calculus, complex analysis, number theory created the definition of limit we use today but didn’t understand it
  • 28. Outline The Concept of Limit Heuristics Errors and tolerances Pathologies Limit Laws Easy laws The direct substitution property Limits by algebra Two more limit theorems
  • 29. Limit Laws Suppose that c is a constant and the limits lim f (x) and lim g (x) x→a x→a exist. Then 1. lim [f (x) + g (x)] = lim f (x) + lim g (x) x→a x→a x→a 2. lim [f (x) − g (x)] = lim f (x) − lim g (x) x→a x→a x→a 3. lim [cf (x)] = c lim f (x) x→a x→a 4. lim [f (x)g (x)] = lim f (x) · lim g (x) x→a x→a x→a
  • 30. Limit Laws, continued f (x) lim f (x) 5. lim = x→a , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a
  • 31. Limit Laws, continued f (x) lim f (x) 5. lim = x→a , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) x→a x→a
  • 32. Limit Laws, continued f (x) lim f (x) 5. lim = x→a , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a
  • 33. Limit Laws, continued f (x) lim f (x) 5. lim = x→a , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a 7. lim c = c x→a 8. lim x = a x→a
  • 34. Limit Laws, continued f (x) lim f (x) 5. lim = x→a , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a 7. lim c = c x→a 8. lim x = a x→a 9. lim x n = an x→a √ √ 10. lim n x = n a x→a
  • 35. Limit Laws, continued f (x) lim f (x) 5. lim = x→a , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a 7. lim c = c x→a 8. lim x = a x→a 9. lim x n = an (follows from 6 and 8) x→a √ √ 10. lim n x = n a x→a
  • 36. Limit Laws, continued f (x) lim f (x) 5. lim = x→a , if lim g (x) = 0. x→a g (x) lim g (x) x→a x→a n n 6. lim [f (x)] = lim f (x) (follows from 3 repeatedly) x→a x→a 7. lim c = c x→a 8. lim x = a x→a 9. lim x n = an (follows from 6 and 8) x→a √ √ 10. lim n x = n a x→a n 11. lim f (x) = n lim f (x) (If n is even, we must additionally x→a x→a assume that lim f (x) > 0) x→a
  • 37. Direct Substitution Property Theorem (The Direct Substitution Property) If f is a polynomial or a rational function and a is in the domain of f , then lim f (x) = f (a) x→a
  • 38. Limits do not see the point! (in a good way) Theorem If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L. x→a x→a
  • 39. Limits do not see the point! (in a good way) Theorem If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L. x→a x→a Example x 2 + 2x + 1 Find lim , if it exists. x→−1 x +1
  • 40. Limits do not see the point! (in a good way) Theorem If f (x) = g (x) when x = a, and lim g (x) = L, then lim f (x) = L. x→a x→a Example x 2 + 2x + 1 Find lim , if it exists. x→−1 x +1 Solution x 2 + 2x + 1 Since = x + 1 whenever x = −1, and since x +1 x 2 + 2x + 1 lim x + 1 = 0, we have lim = 0. x→−1 x→−1 x +1
  • 41. Finding limits by algebraic manipulations Example √ x −2 Find lim . x→4 x −4
  • 42. Finding limits by algebraic manipulations Example √ x −2 Find lim . x→4 x −4 Solution √ √ √ 2 Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2).
  • 43. Finding limits by algebraic manipulations Example √ x −2 Find lim . x→4 x −4 Solution √ 2 √ √ Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). So √ √ x −2 x −2 lim = lim √ √ x→2 x − 4 x→2 ( x − 2)( x + 2) 1 1 = lim √ = x→2 x +2 4
  • 44. Finding limits by algebraic manipulations Example √ x −2 Find lim . x→4 x −4 Solution √ 2 √ √ Write the denominator as x − 4 = x − 4 = ( x − 2)( x + 2). So √ √ x −2 x −2 lim = lim √ √ x→2 x − 4 x→2 ( x − 2)( x + 2) 1 1 = lim √ = x→2 x +2 4 Example√ √ 3 x− 3a Try lim . x→a x −a
  • 45. Two More Important Limit Theorems Theorem If f (x) ≤ g (x) when x is near a (except possibly at a), then lim f (x) ≤ lim g (x) x→a x→a (as usual, provided these limits exist). Theorem (The Squeeze/Sandwich/Pinching Theorem) If f (x) ≤ g (x) ≤ h(x) when x is near a (as usual, except possibly at a), and lim f (x) = lim h(x) = L, x→a x→a then lim g (x) = L. x→a
  • 46. We can use the Squeeze Theorem to make complicated limits simple.
  • 47. We can use the Squeeze Theorem to make complicated limits simple. Example 1 Show that lim x 2 sin = 0. x→0 x
  • 48. We can use the Squeeze Theorem to make complicated limits simple. Example 1 Show that lim x 2 sin = 0. x→0 x Solution We have for all x, 1 −x 2 ≤ x 2 sin ≤ x2 x The left and right sides go to zero as x → 0.