Prove that $\lim_{x\rightarrow +\infty}\left(g(x+\sqrt{x})-g(x)\right)=0$ Let $g:(1,+\infty)\rightarrow \mathbb{R}$, differentiable, such that $|g'(x)|\leq \frac{1}{x}$ for all $x>1$. Show that
$$\lim_{x\rightarrow +\infty}\left(g(x+\sqrt{x})-g(x)\right)=0.
$$
Seems like the mean value theorem is useful. However, not sure how to prove the fact.
Remark:
The hints says that only proving 
$$\lim_{x\rightarrow +\infty}|g(x+\sqrt{x})-g(x)|=0.
$$
won't suffice.

Hint: MVT implies the existence of  $y\in (x,x+\sqrt x)$ such that $|g(x+\sqrt x)-g(x)|=\sqrt x|g'(y)|\leq \sqrt x|1/y|\leq {1\over {\sqrt x}}$.
