Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a periodic continuous function of period $T > 0$, that is $f(x+T)=f(x)$ holds for any $x\in\mathbb{R}$. Show that
$$\lim_{x\to\infty}\frac{1}{x}\int_0^xf(t)dt=\frac{1}{T}\int_0^Tf(t)dt$$
For any $x>0$, there exist a unique pair of $k > 0$ and $0\le a < T$ such that $x=kT+a$. As $x\to\infty$, so will $k\to\infty$. It follwos
$$\begin{align*} \frac{1}{x}\int_0^{x}f(t)dt &=\frac{1}{kT+a}\int_0^{kT+a}f(t)dt\\&=\frac{1}{kT+a}\left(\int_0^{kT}f(t)dt + \int_{kT}^{kT+a}f(t)dt\right) \end{align*}$$
Because $f(t)$ is periodic, thus
$$\int_0^{kT}f(t)dt = k\int_0^Tf(t)dt\qquad\text{and}\qquad\int_{kT}^{kT+a}f(t)dt=\int_{0}^{a}f(t)dt $$
Hence,
$$\frac{1}{x}\int_0^xf(t)dt=\frac{k}{kT+a}\int_{0}^Tf(t)dt + \frac{1}{kT+a}\int_0^af(t)dt$$
Making $k\to\infty$ yields the desired result.