# Proving limit of $f(x+1)-f(x)=0$ for a function $f$ such that $f'(x)\to0$

As the title says i want to prove for a function: $$f: \mathbb{R}\rightarrow\mathbb{R} \lim_{x \to \infty}f'(x)=0$$ Function is also differentiable. So i want to prove that this limit is true: $$\lim_{x\to \infty}(f(x+1)-f(x))=0$$ Probably i would need to use a derivative, but i don't know how to apply it in this way. I was trying to use derivative in a way like this:$\lim_{x\to \infty}(f'(x+1)-f'(x))$ but it didn't help much.

So i got this:$\lim_{x\to \infty}(f'(x+1))$ But how can i calculate that part? Or does it mean it's also equal to $0$, so i already successfully proved it?

Any help or explanation would be appreciated.

Using Mean value theorem there is $x \lt c_x \lt x +1$ such that $f(x+1) - f(x) = f'(c_x) \tag 1$
(1) shows the limit $$\lim_{x\to \infty}(f(x+1)-f(x))$$ exists and by making $x\to\infty$ in (1) we get the result.