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Differential function for which limit as x-> infinity

  1. Dec 20, 2003 #1
    Suppose f is a differential function for which limit as x-> infinity f(x) and limit x->infinity f'(x) both exists and are finite. Which of the following must be true?

    A. limit x-> infinity f'(x) = 0.
    B. limit x0> infinity f''(x) = 0
    C. limit x-> infinity f'(x) = limit x-> infinity f'(x)
    D. f is a constant function
    E. f' is a constant function


    The answer is A. Why are the others wrong (especially explaining why B. is wrong)? Can you provide more formal reason other than it is just intuitive?

    Thanks!!
     
  2. jcsd
  3. Dec 20, 2003 #2

    uart

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    - If A is true then B is also true.

    - Item C is trivially true. (I think you made a typo in C BTW).

    - Both D and E are easily proven false by counter-example.
     
  4. Dec 20, 2003 #3

    uart

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    Unfortunately with this type of question you can't always assume that just because one answer is "the" correct one that the others are necessarily false statements. Sometimes one of the serveral true statements is merely deemed to be "the" correct one by the examiner because it is more fundamental than the others.

    Take the example of statements A and B above. It is easy to show that if A is true then B is also true (that is, A implies B). Note however B does not neccessarily imply A. So, while they are both true, A is a stronger statement than B.
     
  5. Dec 20, 2003 #4

    Hurkyl

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    [itex]\lim_{x \rightarrow \infty} f''(x)[/itex] doesn't necessarily exist.
     
  6. Dec 20, 2003 #5
    I did make a typo in C. It is supposed to read:
    C. limit x-> infinity f(x) = limit x-> infinity f'(x).

    If A is true, B is not necessarily true. The actual answer is A (only). I understand why C-E are false. Why is B.) not necessarily true?

    Thanks so much!!
     
  7. Dec 20, 2003 #6

    Hurkyl

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    Not every differentiable function is twice differentiable; there is no reason to think [itex]f''(x)[/itex] even exists, let alone has a limit as x approaches infinity!

    If you want an explicit example, here's a hint on how to construct one: A simple way for a function not to have a limit at infinity is if it alternates between 2 and -2 infinitely often. The parabolas [itex]y=x^2[/itex] and [itex]y=-x^2[/itex] have second derivative 2 and -2, so the question is can you figure out how to make a curve sewed together from pieces of these two parabolas such that the curve has a limit at infinity, the first derivative always exists, and approaches 0 at infinity?
     
  8. Dec 21, 2003 #7

    uart

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    Whoops, I didn't even think about the possible non-existance of f''.


    Ok my mistake, A Implies B is only true if we are allowed to assume that the limit x-> infinity of f''(x) exists. In that case just let g(x)=f'(x) and g'(x)=f''(x) and you can see that statement B is just a slightly less general version of statement A, but referring to g(x) instead of f(x).
     
    Last edited: Dec 21, 2003
  9. Dec 22, 2003 #8

    uart

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    Ok, I just tried to follow Hurkyl's advice on how to build a function for which limit, x-> infinity, of f'' doesn't exist even though the funtion and first derivative have limits that go to zero.

    The following is what I came up with. Does it look ok to you Hurkyl ?

    Define the sequence s_k as,
    [tex]s_k = 1/k + 2 \sum_{m=1}^{k-1} 1/m [/tex]

    Define the partial function P_k as,

    [tex]p_k(x) = (2 {\rm odd}(k) - 1)((x - s_k)^2 - 1/k^2 )\ :\ s_k - (1/k) \le x < s_k + (1/k) [/tex]

    Note: p_k(x) = 0 : otherwise.

    Then an example of a funtion for which the limit, x->infinity, of f'' does not exist is,

    [tex] f(x) = \sum_{k=1}^{\infty} p_k(x) [/tex].
     
    Last edited: Dec 22, 2003
  10. Dec 22, 2003 #9

    Hurkyl

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    Does [itex]f'(x)[/itex] exist everywhere, though?


    P.S. [itex](-1)^k[/itex] is, IMHO, much clearer than [itex]2 \mathrm{odd}(k) - 1[/itex]
     
  11. Dec 22, 2003 #10

    NateTG

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    Here's a nicer example:
    If
    [tex]f'(x)=e^{-x}sin(e^x)[/tex]
    then the limit for [tex]f'(x)[/tex] is clearly 0.

    [tex]f(x)=-e^{-x}sin(e^x)+\int cos (e^x) dx[/tex]
    Which has a limit as [tex]x \rightarrow \infty[/tex] so that we can make the limit 0 by using the correct constant of integration.

    On the other hand,
    [tex]f''(x)=cos(e^x)-e^{-x}cos(e^x)[/tex]
    clearly has no limit as [tex]x \rightarrow +\infty[/tex] since the first term ocillates with amplitude 1 while the second dissaprears.
     
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