For all $n\in\mathbb{N}$, let $f_n:(0,\infty)\to\mathbb{R}$ be the function defined by $f_n(x)=\frac{n\sin(x/n)}{x(x^2+1)}$. Find the pointwise limit of $(f_n)$.
I know that for all $x>0$ we have that $$|f_n(x)| = \left| \frac{n\sin(x/n)}{x(x^2+1)} \right| \leq \frac{n\cdot (x/n)}{x(x^2+1)} = \frac{1}{x^2+1}.$$ I guess that $\frac{1}{x^2+1}$ is the function that we are looking for. However, I don't know how to prove it by using the definition of pointwise limit.
The pointwise limit $f$ of the sequence $\{f_n\}_{n\in\mathbb{Z}^+}$ you get by computing, for every $x>0$, the limit
$$f(x)=\lim_{n\to\infty} f_n(x)=\lim_{n\to\infty} \frac{n\sin(x/n)}{x(x^2+1)}.$$
Notice that (substituting $h=\frac{1}{n}$) this is the same limit as
$$f(x)=\lim_{h\to0}\frac{\sin(xh)}{xh}\cdot\frac{1}{x^2+1}=\frac{1}{x^2+1},$$
and, so as you can see, we get the same limit that you computed.