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:

HallsofIvy
Homework Helper
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.

quasar987