Is $f(x) = \frac{2x}{1+2x^2}$ uniformly continuous on $\mathbb{R}$? Is $f(x) = \frac{2x}{1+2x^2}$ uniformly continuous on $\mathbb{R}$? Please give your explanation.
I've tried this and got stuck
\begin{align}|f(x) - f(u)|
&= \Big|\frac{2x}{1+2x^2} - \frac{2u}{1+2u^2}\Big| \\
&=\Big|\frac{(-4xu+2)(x-u)}{(1+2x^2)(1+2u^2)}\Big| \\
& = \Big|\frac{-4xu+2}{(1+2x^2)(1+2u^2)}\Big| |x-u|  \\
&\leq \Big(\frac{|-4xu|}{(1+2x^2)(1+2u^2)} + \frac{2}{(1+2x^2)(1+2u^2)} \Big) |x-u| \\
& \leq \Big(\frac{|-4xu|}{(1+2x^2)(1+2u^2)} + \frac{2}{1+2u^2} \Big) |x-u|  \\
&\leq \Big(\frac{|-4xu|}{(1+2x^2)(1+2u^2)} + 2\Big) |x-u|.\end{align}
So I confused to make $$\frac{|-4xu|}{(1+2x^2)(1+2u^2)}$$ to be less than some positive number. What should I do next?
 A: The function $f(x)$ is uniformly continuous on $\mathbb{R}$, Just notice $$\lim_{|x|\to\infty}{f'(x)}=0$$ So the derivative function $f'(x)$ is bounded on $\mathbb{R}$, we suppose that $M>0$ satisfies$|f'(x)|<M$ for all real $x\in R$,
so for ${\forall} \epsilon>0$, and ${\forall}$ $x_{1}$<$x_{2}$ satisfies$|x_{1}-x_{2}|$<$\frac{\epsilon}{M}$, by Lagrange Mean Value Theorem ,$\exists x_{1}<x_{0}<x_{2}$ satisfy $f(x_{1})-f(x_{2})=f'(x_{0})(x_{2}-x_{1})$, so $|f(x_{1})−f(x_{2})|=|f'(x_{0})|(x_{2}-x_{1})\leq M(x_{2}-x_{1})=\epsilon$. We have  $f(x)$ is uniformly continuous. More, $f(x)$ is even the lipschitz function.
A: Finish what you started using
$$\frac{4|xu|}{(1+2x^2)(1+2u^2)}= \frac{4|xu|}{1+2x^2+2u^2 + 4x^2u^2} \leqslant \frac{2|x||u|}{x^2 + u^2} \leqslant 1$$
The last inequality follows from $(|x| - |u|)^2 \geqslant 0$.
A: Actually any continuous function $f$ on $\mathbb R$ such that $\lim_{x\to \pm\infty}f(x)=0$ is uniformly continuous on $\mathbb R.$ We have that hypothesis satisfied with $f(x)=2x/(1+2x^2).$
