Evalute $\lim\limits_{n\to\infty}\int _{0}^{1}\frac{nf(x)}{1+n^2x^2}\,dx$ Suppose $f\colon [0, 1] \to \mathbb{R}$ is a continuous function, Find the following limit:
$$\lim\limits_{n\to\infty}\int _{0}^{1}\frac{nf(x)}{1+n^2x^2}\,dx$$
I know the  answer is equal   $\frac{\pi}{2}f(0) $, but I don't prove that.
Any ideas or insight would be greatly appreciated.
 A: With the change $y=nx$ you get $$\int _{0}^{1}\frac{nf(x)}{1+n^2x^2}\,dx = \int _{0}^{n}\frac{f(\frac{y}{n})}{1+y^2}\,dx.$$ Now Lebesgue theorem should allow you to conclude.
A: I get that the limit is
$\frac12 (1+\pi \coth(\pi))f(0)
$.
Since this disagrees with
what is presumably the correct answer,
I would like to know
where and what my error is.
Thanks.
$\begin{array}\\
\lim\limits_{n\to\infty}\int _{0}^{1}\frac{nf(x)}{1+n^2x^2}\,dx
&=\lim\limits_{n\to\infty}\frac1{n}\sum_{k=0}^{n-1}\frac{nf(k/n)}{1+n^2(k/n)^2}
\qquad\text{(the error is probably here, but I'm not sure why)}\\
&=\lim\limits_{n\to\infty}\sum_{k=0}^{n-1}\frac{f(k/n)}{1+k^2}\\
&=\lim\limits_{n\to\infty}\sum_{0\le k\lt n^a}\frac{f(k/n)}{1+k^2}
+\lim\limits_{n\to\infty}\sum_{n^a\ge k\le n-1}\frac{f(k/n)}{1+k^2}
\qquad\text{where }\frac12 < a <1\\
&=\lim\limits_{n\to\infty}\sum_{0\le k\lt n^a}\frac{f(k/n)}{1+k^2}
\qquad\text{since }\lim\limits_{n\to\infty}\sum_{n^a\le k\le n-1}\frac{f(k/n)}{1+k^2}=0\\
&=f(0)\lim\limits_{n\to\infty}\sum_{0\le k\lt n^a}\frac{1}{1+k^2}\\
&=f(0)\sum_{k=0}^{\infty}\frac{1}{1+k^2}\\
&=\frac12 (1+\pi \coth(\pi))f(0)
\qquad\text{(according to Wolfy)}\\
&\approx 2.076674f(0)
\end{array}
$
