Prove that it is not uniformly convergent $(0,\infty)$ \begin{align} f_n(x)=\frac{nx}{1+n^2x^2} \end{align} I know how to solve this question in another way. I only wonder whether the solution can be done in the following way.
We have $f_n(x) \to 0$. If it is not uniformly convergent, n depends on $(x,\epsilon)$. If it is convergent, then $\forall \epsilon \quad \exists N, n\geq N$ such that $|f_n(x)-0| < \epsilon$ \begin{align} |f_n(x)-0|=\frac{nx}{1+n^2x^2} < \frac{1}{nx} < \frac{1}{Nx} \end{align} Then if we fix $\epsilon$, we can choose $\epsilon =\frac{1}{Nx} \to N=\frac{1}{\epsilon x}$.
Since N depends on $x$, it is pointwise convergent. Is that solution true ? Thank you for your help.
Yes, you proved that $$\forall x>0 \; \forall \epsilon>0 \; \exists N=1/x\epsilon \; :$$
$$n>N\implies |f_n (x)-0|<\epsilon $$ this is the pointwise convergence to zero function at $(0,+\infty) $.
for uniform convergence, observe that for $n $ great enough, $$\sup_{x>0}|f_n (x)-0|\ge |f_n (1/n)-0|=\frac {1}{2} .$$