# Limit of $f(x+\sqrt x)-f(x)$ as $x \to\infty$ if $|f'(x)|\le 1/x$ for $x>1$

Given a differentiable function $f:(1,+\infty) \to \mathbb{R}$, and
$|f'(x)| \le 1/x$ for every $x>1$. Prove that:

$$\lim_{x\to +\infty} \left(f(x+\sqrt{x})-f(x) \right)=0.$$

#### Solutions Collecting From Web of "Limit of $f(x+\sqrt x)-f(x)$ as $x \to\infty$ if $|f'(x)|\le 1/x$ for $x>1$"

By showing that
\begin{eqnarray}
&&\frac{\left| f(x+\sqrt{x})-f(x) \right|}{(x+\sqrt{x})-x} \le \mathop{\max} \limits_{x\le\theta\le x+\sqrt{x}} \left| f'(\theta) \right| = \frac{1}{x}\\
&\Rightarrow& \left| f(x+\sqrt{x})-f(x) \right| \le \frac{1}{\sqrt{x}} \\
&\Rightarrow& 0\le \mathop {\lim }\limits_{x \to \infty } \left| {f(x+\sqrt{x})-f(x)} \right| \le \mathop {\lim }\limits_{x \to \infty } \frac{1}{\sqrt{x} } = 0 \\
&\Rightarrow& \mathop {\lim }\limits_{x \to \infty } \left| {f(x+\sqrt{x})-f(x)} \right| = 0
\end{eqnarray}
clearly we have
\begin{eqnarray}
0 &=& \mathop {\lim }\limits_{x \to \infty } -\left| {f(x+\sqrt{x})-f(x)} \right| \\
&\le&\mathop {\lim }\limits_{x \to \infty } f(x+\sqrt{x})-f(x) \\
&\le& \mathop {\lim }\limits_{x \to \infty } \left| {f(x+\sqrt{x})-f(x)} \right| \\
&=& 0 \\
\Rightarrow && \mathop {\lim }\limits_{x \to \infty } {f(x+\sqrt{x})-f(x)} = 0
\end{eqnarray}

This is bit strange as earlier provided answers have ignored the inequality $|f'(x)| \leq 1/x$ altogether and confused it with $|f(x)| \leq 1/x$ and surprisingly they have been given upvotes. I believe there might have been a typo in the question which has now been fixed.

The correct way to solve the problem is through mean value theorem. We have $$|f(x + \sqrt{x}) – f(x)| = |\sqrt{x}f'(c)|$$ for some $c \in (x, x + \sqrt{x})$. Now we know that $|f'(c)| \leq 1/c < 1/x$ hence we see that $$0 \leq |f(x + \sqrt{x}) – f(x)| < \sqrt{x}/x = 1/\sqrt{x}$$ Clearly by Sanwich theorem (aka Squeeze theorem) we get $$\lim_{x \to \infty}|f(x + \sqrt{x}) – f(x)| = 0$$ and since $$-|f(x + \sqrt{x}) – f(x)| \leq f(x + \sqrt{x}) – f(x) \leq |f(x + \sqrt{x}) – f(x)|$$ applying Sandwich theorem again we get $$\lim_{x \to \infty}f(x + \sqrt{x}) – f(x) = 0$$

Edit: My answer is for the originally posted question, where it said $|f(x)|<\frac1 x$.

In this case, the assumption that $f$ is differentiable is actually not necessary.

As $f(x)$ is bounded for any $x\in(1,\infty)$, you can just calculate
$$\left|f\left(x+\sqrt{x}\right)-f(x)\right|\leq\left|f\left(x+\sqrt{x}\right)\right|+\left|f(x)\right|\leq\frac{1}{x+\sqrt{x}}+\frac1 x\leq\frac2 x\,.$$
This implies $$0\leq\lim\limits_{x\to\infty}\;\left|f\left(x+\sqrt{x}\right)-f(x)\right|\leq\lim\limits_{x\to\infty}\;\frac2 x\leq 0\,.$$