Let $f:[0,\infty)\rightarrow R$ be differentiable on $(0,\infty)$ and assume that $f'(x)\rightarrow b$ as $x\rightarrow\infty$.
(a) Show that for any $h>0$, we have $\lim_{x\rightarrow\infty}\frac{f(x+h)-f(x)}{h}=b$.
(b) Show that if $f(x)\rightarrow a$ as $x\rightarrow\infty$, then $b=0$.
(c) Show that $\lim_{x\rightarrow\infty}\frac{f(x)}{x}=b$.
Note: this is exercise 6.2.16 from "Introduction to Real Analysis" by Bartle & Sherbert, that I'm using for self-study. Part (a) is an easy one, but am stuck with parts (b) and (c). My solution to part (a) would be as follows:
By definition, $\lim_{x\rightarrow\infty}f'(x)=b$ means that for all $\epsilon>0$ there exists $K(\epsilon)$ such that if $x>K$, then $|f'(x)-b|<\epsilon$. Now, for given $h>0$, it is that $x+h>K$ too, and by MVT there exists $c$ such that $x<c<x+h$ and $f'(c)=\frac{f(x+h)-f(x)}{h}$. As $c>K$, then $\lvert f'(c)-b \rvert <\epsilon$ too, and thus $\left|\frac{f(x+h)-f(x)}{h}-b\right|<\epsilon$ for all $x>K$; which, by definition, means that $\lim_{x\rightarrow\infty}\frac{f(x+h)-f(x)}{h}=b$.
Another note: Bartle & Sherbert book has no solutions for all exercises, but for many of them there exist publicly available solutions. For this one, however, the only solution available is here. However, arguments provided there to prove parts (b) and (c) just don't seem rigorous enough to me.
