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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.

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

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By Lagrange mean value theorem, $$|g(x + \sqrt{x}) - g(x)| = |g'(\xi)(x + \sqrt{x} - x)| \leq \frac{1}{\xi}\sqrt{x} \leq \frac{\sqrt{x}}{x} \to 0$$ as $x \to +\infty$. Here $\xi \in (x, x + \sqrt{x})$ so that the second inequality holds.

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