If $f_n (x)=\frac{n \sin x}{x (1+n^2 x^2)}$ then evaluate limit of integration $f_n(x)$ over $0 \to 1$ as $n \to \infty$ In this problem,
I tried to dominated convergence theorem but I couldn't get any dominated function. How to find limit of this integration?
Any hints or comments are welcomed.
 A: I am assuming that you have
$$
f_n(x)=\frac{\sin(x)}{x}\frac{n}{1+n^2x^2}
$$
and want
$$
\lim_{n\to\infty}\int_0^1f_n(x)\,\mathrm{d}x
$$
First note that
$$
\begin{align}
\lim_{n\to\infty}\int_0^1\frac{n}{1+n^2x^2}\,\mathrm{d}x
&=\lim_{n\to\infty}\int_0^n\frac1{1+x^2}\,\mathrm{d}x\\
&=\frac\pi2
\end{align}
$$
For $n\gt\frac1x$, $\frac{n}{1+n^2x^2}$ tends monotonically to $0$. Thus, using Dominated convergence on intervals avoiding $0$, we can deduce that
$$
\lim_{n\to\infty}\int_0^1g(x)\frac{n}{1+n^2x^2}\,\mathrm{d}x=\frac\pi2\lim_{x\to0}g(x)
$$
for any $g$ continuous on $[0,1]$.
For this question, $g(x)=\frac{\sin(x)}{x}$.
A: We have for $f_n(x)=\frac{n}{1+n^2x^2}$ $\lim_{n\to \infty}\frac{n}{1+n^2x^2}=0$ for $x\ne 0$ and $\lim_{n\to \infty}\frac{n}{1+n^2x^2}=\infty$ for $x=0$.  
Heuristically, then we anticipate that for an adequately smooth function $\phi$ we have
$$\begin{align}
\lim_{n\to \infty}\int_0^1  f_n(x)\phi(x)dx&=\phi(0)\lim_{n\to \infty}\int_0^1  f_n(x)\,dx\\\\
=\frac{\pi}{2}\phi(0)
\end{align}$$
To make this more rigorous, let's assume that $\phi$ is continuous with $|\phi(x)|\le M$ on $[0,1]$.  Now, we examine the integral
$$\begin{align}
\left|\int_0^1 f_n(x)\,(\phi(x)-\phi(0))\,dx\right| & \le \int_0^1 f_n(x)\,\left|\phi(x)-\phi(0)\right|\,dx \tag 1\\\\
&=\int_0^{\delta} f_n(x)\,|\phi(x)-\phi(0)|\,dx\\\\
&+\int_{\delta}^1 f_n(x)\,|\phi(x)-\phi(0)|\,dx \tag 2
\end{align}$$
for $\delta>0$, where we simply split the integral in going from $(1)$ to $(2)$.
Given $\epsilon>0$, we choose $\delta(\epsilon)$ so that $|\phi(x)-\phi(0)|<\epsilon/\pi$ for $|x|<\delta(\epsilon)$.  Thus, we have for the first integral on the right-hand side on $(2)$
$$\begin{align}
\int_0^{\delta} f_n(x)\,|\phi(x)-\phi(0)|\,dx & \le \frac{\epsilon}{\pi}\int_0^{\delta} f_n(x)\,dx\\\\
&=\frac{\epsilon}{\pi}\,\arctan(n\delta)\\\\
&\le\frac{\epsilon}{2}
\end{align}$$
We have for the second integral on the right-hand side of $(2)$
$$\begin{align}
\int_{\delta}^{1} f_n(x)\,|\phi(x)-\phi(0)|\,dx & \le 2M \int_{\delta}^{1} f_n(x)\,dx\\\\
&\le 2M \frac{n}{1+n^2\delta^2}\\\\
&\le\frac{2M}{n\delta^2}\\\\
&\le \frac{\epsilon}{2}
\end{align}$$
whenever $n>N(\epsilon)=\frac{4M}{\delta^2\,\epsilon}$.
Putting it all together, we have that for any given $\epsilon>0$, there exists a $\delta(\epsilon)>0$, and an $N(\epsilon)=\frac{4M}{\delta^2(\epsilon)\,\epsilon}$ such that 
$$\left|\int_0^1 f_n(x)\,(\phi(x)-\phi(0))\,dx\right|<\epsilon$$
whenever $n>N$, which yields the desired result.  We simply take $\phi(x)=\sin x/x$, for $x \ne 0$ and $\phi(0)=1$.
