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Question about limits

  1. Sep 1, 2012 #1
    suppose there is a function f(x), and it's limit as x goes to infinity is 0.

    Is there a theorem that says it's derivative, f'(x), also approaches 0 as x goes to infinity?

    Thanks.
     
  2. jcsd
  3. Sep 1, 2012 #2

    micromass

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    Probably not, since it is not true. Consider

    [tex]f(x)=\frac{\sin(x^2)}{x}[/tex]

    What is true, that if [itex]f^\prime(x)[/itex] has a limit, then this limit must be 0.
     
  4. Sep 1, 2012 #3

    Bacle2

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    If f is differentiable in (-oo,oo) , use the mean value theorem:

    f(b)-f(a)=f'(c)(b-a) . Maybe you can partition [0,oo)into [0,1],[0,2],...,[n,n+1].

    Then:

    f(1)=f(0)+f'(co) ;f(2)=f(1)+f'(c1) ;.....f(n)=f(n-1)+f'(c_(n-1)).

    Then you can find a closed form for f(n).
     
  5. Sep 1, 2012 #4

    haruspex

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    micromass's example shows this isn't going to work.
     
  6. Sep 1, 2012 #5

    Bacle2

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    Of course I'm assuming f'(x) is defined as x-->oo , that is implied in my argument.

    Basically, take [0,b] . Then

    f(b)-f(0)=f'(c)(b)

    If f' is defined everywhere and we let b-->oo , then the limit cannot be 0 unless f'(c) decreases to zero.

    If the OP says "its derivative approaches 0 as x --> infinity" seems to me to assume that the derivative is defined as x-->oo.
     
    Last edited: Sep 1, 2012
  7. Sep 1, 2012 #6

    haruspex

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    You mean, that the limit is defined? If your argument relies on that then it needs to be stated.
     
  8. Sep 1, 2012 #7

    Bacle2

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    From the OP:" it's derivative, f'(x), also approaches 0 as x goes to infinity? "


    This looks to me like an assumption that f'(x) is defined as x-->oo
     
  9. Sep 1, 2012 #8

    Bacle2

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    Maybe the OP can clarify the conditions of the problem to eliminate ambiguity?
     
  10. Sep 1, 2012 #9

    haruspex

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    No, that's what he's trying to prove. He didn't say "if f' approaches a limit, that limit is 0", so the most reasonable interpretation is that he wants to prove "f' approaches a limit, and that limit is 0".
     
  11. Sep 2, 2012 #10

    Bacle2

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    It's not clear to me either way. f'(x) is said to exist without any qualification; I see

    no reason to assume it exists in a specific subset of the real line only, nor reason

    to assume otherwise. In your interpretation, why didn't the OP say something like

    is f'(x) defined, and if so , what is its limit. He refers to f'(x) which states that f'(x)

    exists. It may exist somewhere or everywhere.

    The problem is posed sloppily ; I think out of basic manners, the OP should clarify.
     
  12. Sep 2, 2012 #11

    pwsnafu

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    Then it does not follow that the limit ##\lim_{x\rightarrow\infty}f'(x)## is defined. The statement "f' exists" is limited to ##x \in \mathbb{R}## because the domain of f is the real numbers and not the extended reals. What happens as ##x\rightarrow\infty## is considered a separate condition, and must explicitly be mentioned.
     
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