I have this theorem in solution manual to Rudin PMA:
Theorem: Suppose $f$ is a function with domain $(0,\infty).$ If $f(x)\to L$ as $x\to\infty$ and $f'(x)$ is uniformly continuous on $(0,\infty)$, then $f'(x)\to0$ as $x\to\infty$.
Proof: For, if not, let $x_n\to\infty$ be a sequence such that $f'(x_n)\ge\epsilon>0$ for all $n$. (We can assume $f'(x_n)$ is positive by replacing $f$ with $-f$ if necessary.) Let $\delta$ be such that $|f'(x)-f'(y)|<\dfrac{\epsilon}2$ if $|x-y|<\delta$. We then have $f'(y)>\dfrac{\epsilon}2$ if $|x_n-y|<\delta$, and so $$|f(x_n+\delta)-f(x_n-\delta)|\ge 2\delta\cdot\dfrac{\epsilon}{2}=\delta\epsilon.$$ And proof goes on ...
I can't understand how we got $|f(x_n+\delta)-f(x_n-\delta)|\ge 2\delta\cdot\dfrac{\epsilon}{2}.$
By the mean value theorem,\begin{align}\bigl|f(x_n+\delta)-f(x_n-\delta)\bigr|&=\bigl|f'(c)\bigr|.\bigl((x_n+\delta)-(x_n-\delta)\bigr)\\&=\bigl|f'(c)\bigr|.2\delta,\end{align}for some $c\in(x_n-\delta,x_n+\delta)$.