<s>[INST] 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?
 [/INST] 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 $(|x| - |u|)^2 \geqslant 0$.
</s>