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Analysis differentiability and limits - please help

  1. May 17, 2009 #1

    C.E

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    1. Suppose that f is differentiable on R and [tex]\lim_{x \rightarrow \infty} f'(x) [/tex] = M. Show that [tex]\lim_{x \rightarrow \infty} (f(x+1)-f(x)) [/tex] also exists, and compute it.

    3. I am pretty sure the limit will be equal to m. Here is my attempt.

    [tex]\lim_{x \rightarrow \infty} f'(x) [/tex] = [tex]\lim_{x \rightarrow \infty} \lim_{h

    \rightarrow 0} \frac{f(x+h)-f(x)}{h} [/tex]

    By continuity this is equal to:

    [tex]\lim_{h \rightarrow 0} \lim_{x \rightarrow \infty} \frac{f(x+h)-f(x)}{h} [/tex]

    Fixing h=1

    This is equal to [tex]\lim_{h \rightarrow 0} \lim_{x \rightarrow \infty} \frac{f(x+1)-f(x)}{1} [/tex]

    =[tex]\lim_{x \rightarrow \infty} (f(x+1)-f(x)) [/tex].

    Have I done this right? I don't think you can just swap the limits like that can you? Does anybody have any suggestions?
     
    Last edited: May 17, 2009
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  3. May 17, 2009 #2

    HallsofIvy

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    You certainly can't say that you can swap "by continuity"! And how can you "fix" h= 1 and than take the limit as h goes to 0?

    I suggest that you apply the mean value theorem to the interval between x and x+ 1, then take the limit as x goes to infinity.
     
  4. May 17, 2009 #3

    quasar987

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    Yeah, that limit swapping it kinda dubious!

    Here's a neater way. Fix x in R. By the mean value theorem, there exists a number c(x) in [x,x+1] such that f(x+1)-f(x)=f'(c(x)).
     
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