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?

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 $f(x)=2x/(1+2x^2).$
