How to prove $\lim\limits_{n\to\infty}\int^\pi_0 \pi f(nx)g(x)\ \mathrm {d}x = \int^\pi_0 f(x)\ \mathrm {d}x \int^\pi_0 g(x)\ \mathrm {d}x$ Question
Let $f$ be a continuous function on $\mathbb {R}$ such that $f(x + \pi) = f(x)$ for all $x \in \mathbb {R}$. Prove that, if $g$ is a continuous function on $[0, \pi]$, then $$\lim\limits_{n\to\infty}\int^\pi_0 \pi f(nx)g(x)\ \mathrm {d}x = \int^\pi_0 f(x)\ \mathrm {d}x \int^\pi_0 g(x)\ \mathrm {d}x.$$

This question just appeared in my final calculus examination this afternoon and I could not do it ):. I am just really bad at proofs. However, on reading the question, the condition of $f(x + \pi) = f(x)$ on $f$ tells me that $f$ must be some kind of periodic function and I see $\pi$ all around, so I am guessing we will have to make use of some trigonometry. Any intuitive explanations as to how the equality can be proven will be greatly appreciated :)
 A: Let $\epsilon>0$. Choose $N$ so that whenever $|x-y|<\pi/N$, we have $|f(nx)g(x)-f(nx)g(y)|<\epsilon/\pi$. This can be done since both $f$ and $g$ are uniformly continuous functions.
Now, for $n\ge N$
$$\eqalign{
\int_0^\pi \pi f(nx)g(x)\,dx&= \sum_{m=1}^n \int_{(m-1)\pi/n}^{m\pi/n}\pi f(nx)g(x)\,dx\cr
&\le\sum_{m=1}^n \Bigl[{\epsilon\over\pi}\cdot{\pi\over n}
+\int_{(m-1)\pi/n}^{m\pi/n} \pi f(nx) g(m\pi/n)\,dx\Bigr]\cr
&=\epsilon +\sum_{m=1}^n\Bigl[ g(m\pi/n) \int_{(m-1)\pi}^{m\pi} {\pi\over n}f(u)\,du\Bigr]\cr
&=\epsilon +\sum_{m=1}^n \Bigl[g(m\pi/n)\cdot{\pi\over n} \int_0^\pi f(x)\,dx\Bigr]\cr
&=\epsilon + \int_0^\pi f(x)\, dx \cdot\Bigl[\sum_{m=1}^n g(m\pi/n) \cdot{\pi\over n}\Bigr]
}$$
Since $\epsilon$ is arbitrary,
taking limits of both sides of the above gives
$$\limsup_{n\rightarrow\infty}\int_0^\pi \pi f(nx)g(x)\,dx \le \int_0^\pi f(x)\,dx \cdot\int_0^\pi g(x)\, dx.$$
In a similar manner, ones shows
$$
\liminf_{n\rightarrow\infty}\int_0^\pi \pi f(nx)g(x)\,dx \ge \int_0^\pi f(x)\,dx \cdot\int_0^\pi g(x)\, dx,
$$
whence the result follows.
