Suppose that $f$ is differentiable on $\mathbb{R}$ and $\lim_{x\to \infty}f'(x)=M$. Show that $\lim_{x\to \infty}f(x+1)-f(x)$ exists and find it. I've been stuck on this question for a long time now and was wondering if anyone could show me how it's done. So far I have done the following:
Since $\lim_{x\to \infty}f'(x)=M$ then $\forall \epsilon ,\exists A >0$  s.t. if $x>A$ then $|f'(x)-M|<\epsilon$. So we have that $M-\epsilon<f'(x)<M+\epsilon$. By MVT on $[x,x+1]$ for $x>A$, $\exists c\in (x,x+1)$ s.t. ${f(x+1)-f(x)\over 1}=f'(c)$. Therefore, $M-\epsilon<f(x+1)-f(x)<M+\epsilon$. However, I do not know what to do after this.

After this you have $|f(x+1)-f(x)-M|<\epsilon$ which signifies that the limit exists and equals to $M$, or you can do better, for every $x$ there exists $c_x\in[x+1,x]$ such that:
$$f(x+1)-f(x)=f'(c_x) $$
now because $\lim_{x\to \infty} c_x=+\infty$ (why? because $c_x\geq x+1$)and $f$ continue gives you:
$$\lim_{x\to \infty}f(x+1)-f(x)=\lim_{x\to \infty}f'(c_x)=\lim_{T\to \infty}f'(T)=M $$
