Prove that if $\lim_{x\to \infty} f'(x) = 0$, then $\lim_{x\to \infty} f(x+1)-f(x) = 0$

Question

I'm working on this proof and I think I have a sketch but I'm not sure it's rigorous enough.

Suppose $f:\Bbb R \to \Bbb R$ is differentiable and that$$\lim_{x\to \infty} f'(x) = 0$$ Prove that $$\lim_{x\to \infty} f(x+1)-f(x) = 0$$

Proof:

We want to prove that for any $\epsilon > 0$, there is an $x_0$ such that $|f(x+1)-f(x)|< \epsilon$ for all $x \ge x_0$. By the Mean Value Theorem, there is some $c \in (x, x+1)$ such that $f(x+1) - f(x) = f'(c)$. By the convergence of $f'(x)$, there is some $y_0$ such that $|f'(x)| < \epsilon$ for all $x \ge y_0$. Thus $$x \ge y_0 \implies c>y_0 \implies |f'(c)| = |f(x+1) - f(x)| < \epsilon$$ and $x_0 = y_0$. $\square$

Thanks!

Answer 1

Yes your proof is well rigorous, well done!

Let me just propose a more concise proof that does not require bringing up any $\epsilon$: by the mean value theorem there is a function $\xi: \mathbb R\to \mathbb R$ such that $$f(x+1)-f(x) = f(\xi(x))$$ and $x<\xi(x)< x+1$. Then $$\lim_{x\to\infty} f(x+1)-f(x) = \lim_{x\to\infty} f(\xi(x)) = 0.$$