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.

Here is an easier way out. By Mean Value theorem, given any $x$, there exists $y \in [x,x+1]$ such that
$$f(x+1) - f(x) = \dfrac{f(x+1)-f(x)}{(x+1)-x} = f'(y)$$
Now as $x \to \infty$, since $y \in [x,x+1]$, we have $y \to \infty$. Hence, we obtain that
$$\lim_{x \to \infty} \left(f(x+1) - f(x)\right) = \lim_{y \to \infty} f'(y) = M$$
