Evaluating the limit of the given integral Evaluate the following:
  $\displaystyle\lim_{n\to\infty} \displaystyle\int_0^\infty \dfrac{n\sin x}{1+n^2x^2}\ dx$


Attempt: My idea to use the Dominated Convergence Theorem. It can easily be shown that $f_n(x)=\dfrac{n\sin x}{1+n^2x^2}$ converges pointwise to $0$ on $\left(0,\infty\right)$ but I'm having difficulties on trying to get an integrable function $g$ that would satisfy $|f_n|\leq g$. I tried using the inequalities $|\sin t|\leq t$ for small $t$ and of course the obvious one which is $|\sin t|\leq 1$ but the function I get is not integrable. Any help would be great.
Hint: Let $u=nx$, then your integral becomes
\begin{align}
\int^\infty_0 \frac{\sin\frac{u}{n}}{1+u^2}\ du
\end{align}
This is not a complete answer: 

The integral from Jacky Chong can be evaluated with Mathematica:
$$\int_0^{\infty } \frac{\text{Sin}\left[\frac{u}{n}\right]}{1+u^2} \
\, du=\frac{\sqrt{\pi } \text{MeijerG}\left[\{\{0\},\{\}\},\left\{\{0,0\},\left\{-\frac{1}{2}\right\}\right\},\frac{1}{4n^2}\right]}{4 n}$$
Therefore the solution can be expressed in form of a MeijerG-Function and can  be further analyzed by an asymptotic series expansion. This leads to the same result as from Claude Leibovici.