# Integral dependent only on value at $0$?

I was asked in today's exam to prove/disprove the following: ($f$ is a continuous function in $[0,1]$)

$$\lim_{n \rightarrow \infty} \int_{0}^{1} \left(\frac{n f(x)}{1 + n^2x^2} \right) dx = \frac \pi 2 f(0)$$

I tried out a few functions $f(x)$, and decided that it was probably true. But all I could do was to reduce the integral to

$$\lim_{n \rightarrow \infty} \int_{0}^{1} \left(\frac{n f(x)}{1 + n^2x^2} \right) dx = \lim_{n \rightarrow \infty} \int_{0}^{n} \frac{f(x/n)}{1 + x^2} dx$$

• Extend $f$ to a function defined on $\mathbb R$ by $\tilde f(x)=0$ for $x>1$, then $$\int_0^n \frac{f(x/n)}{1+x^2}dx=\int_0^\infty \frac{\tilde f(x/n)}{1+x^2}dx\,.$$
• Then, for all $x\geq0$, $$\Big|\frac{\tilde f(x/n)}{1+x^2}\Big|\leq\max_{[0,1]}|f(x)|\frac{1}{1+x^2}\,.$$ The function on the right hand side is integrable on $[0,\infty)$.
• Thus you can use the dominated convergence theorem to prove that $$\lim_{n\to\infty}\int_0^n \frac{f(x/n)}{1+x^2}dx=\int_0^\infty \frac{\lim_{n\to\infty}\tilde f(x/n)}{1+x^2}dx=\int_0^\infty \frac{ f(0)}{1+x^2}dx=f(0)\frac{\pi}{2}\,.$$