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Analysis differentiability and limits - please help

  • Thread starter C.E
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  • #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?
 
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Answers and Replies

  • #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.
 
  • #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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