I solved this problem .
Is my answer a correct ?
$$ x \in [0,\infty) ,\lim_{n \to \infty} \frac{nx}{1+n^2x^2}=0\ \ \ $$ Is $ \frac{nx}{1+n^2x^2} $ converged uniformly on $0$ ?
My solution
$$ \left|\frac{nx}{1+n^2x^2}\right|<\left|\frac{nx}{n^2x^2}\right|=\left|\frac{1}{nx}\right|<\epsilon\\ \ ∴\frac{1}{\epsilon|x|}<n $$
$ \frac{nx}{1+n^2x^2} $ is not converged uniformly on $0$.
To prove that $\frac{nx}{1+n^2x^2}$ does not converge uniformly to $0$ on $[0,\infty)$, by definition, we need to find a number $\epsilon>0$ such that for each $n$ $$\left|\frac{nx}{1+n^2x^2}-0\right|\geq \epsilon\mbox{ for some }x\in [0,\infty).$$
Take $\epsilon=\frac{1}{2}$. Then for each $n$, we can take $x=\frac{1}{n}\in [0,\infty)$ which implies that $$\left|\frac{nx}{1+n^2x^2}-0\right|=\left|\frac{n\cdot\frac{1}{n}}{1+n^2\cdot\frac{1}{n^2}}\right|=\frac{1}{2}\geq \epsilon.$$ This proves that $\frac{nx}{1+n^2x^2}$ does not converge uniformly to $0$ on $[0,\infty)$.
we know that $(1-nx)^2\geq0,$ so taht $2nx\leq1+{nx}^2,$ hence $\frac{nx}{1+{nx}^2}\leq\frac{1}{2}$ this shoes that sequence for $\epsilon\leq \frac{1}{2}$ does not converge uniformly to $0$
