I have solved the following exercise and I would like someone to check and tell me if my solution is right. Thank you.
Let $f_n(x)=\frac{nx}{1+nx^2}$
(a) Find the pointwise limit of $(f_n)$ for all $x\in (0,\infty)$
(b) Is the convergence uniform on $(0,\infty)$?
(c) Is the convergence uniform on $(0,1)$?
(d) Is the convergence uniform on $(1,\infty)$?
My solution (NOTE: corrected parts (b) and (c) thanks to the answers below):
(a) $f_n(x)=\frac{nx}{1+nx^2}=\frac{1}{\frac{1}{nx}+x}\xrightarrow[]{n\to\infty} \frac{1}{x} $
In what follows we use the fact that if $f_n \to f$ uniformly on $(0,\infty)$ or $(0,1)$ then $|f_n(a_n)-f(a_n)|\xrightarrow{n\to\infty}0$ (where $(a_n)$ is a sequence of points belonging to these spaces.)
(b) Consider the sequence $(a_n)=\frac{1}{n}$; then $|f_n(a_n)-\frac{1}{a_n}|=|\frac{n}{n+1}-n|=|-\frac{n^2}{n+1}|=\frac{n^2}{n+1}\xrightarrow{n\to\infty}\infty$ thus the convergence is not uniform here.
(c) The convergence is not uniform by the same reasoning as in part (b)
(d) $|f_n(x)-\frac{1}{x}|=\frac{1}{x(1+nx^2)}\leq\frac{1}{1\cdot (1+1\cdot n)}=\frac{1}{1+n}\leq\frac{1}{n}$. Thus, given $\varepsilon > 0$ we can choose $N >\frac{1}{\varepsilon}$ (independent of $x$) and it follows that $n\geq N$ implies $|f_n(x)-\frac{1}{x}|\leq\varepsilon$ for all $x\in (0,1)$ so here we have uniform convergence.
It is almost correct. Here is a problem in parts (b) and (c), which is this: you used the inequality$$\left|f_n(x)-\frac1x\right|\leqslant\frac1{nx^3}$$and then you stated that, since $\frac1{nx^3}$ is not arbitrarily small uniformly, then the convergence is not uniform. That is not correct. For instance, $\frac xn$ converges uniformly to $0$ on $[0,1]$. But you also have $|f_n(x)-0|\leqslant1$ on $[0,1]$, and the constant function $1$ is not arbitrarily small uniformly.
In order to do (b) and (c), just use the fact that$$(\forall n\in\Bbb N):f_n\left(\frac1n\right)=\frac n{n+1}\to_{n\to\infty}1.$$