1. May 17, 2009

### C.E

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

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

$$\lim_{x \rightarrow \infty} f'(x)$$ = $$\lim_{x \rightarrow \infty} \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$

By continuity this is equal to:

$$\lim_{h \rightarrow 0} \lim_{x \rightarrow \infty} \frac{f(x+h)-f(x)}{h}$$

Fixing h=1

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

=$$\lim_{x \rightarrow \infty} (f(x+1)-f(x))$$.

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
2. May 17, 2009

### HallsofIvy

Staff Emeritus
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. May 17, 2009

### quasar987

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)).