Let $f : (1, +\infty) \rightarrow \mathbb{R}$ be differentiable s.t $|f'(x)| ≤ 1/x$ for all $x > 1$. Show that $(f(x + \sqrt{x}) - f(x)) = 0$ Let $f : (1, +\infty) \rightarrow \mathbb{R}$, be differentiable, such that $|f'(x)| ≤ 1/x$ for all $x > 1$. Show that   
$\lim_{x \rightarrow +\infty} (f(x + \sqrt{x}) - f(x)) = 0$
I started with showing that $\lim_{x \rightarrow +\infty} |f(x + \sqrt{x}|=0$. I don't know how to continue, but I suspect this will be a crucial part at the end of the proof

Hint: by the mean value theorem,
$$ \frac{f(x+\sqrt{x})-f(x)}{\sqrt{x}}=f^{\prime}(t) $$
for some $t\in (x,x+\sqrt{x})$.
