If $f : \mathbb{R} \rightarrow \mathbb{R}$ is $T$-periodic then $\lim\limits_{L \rightarrow +\infty} \int_a^b f(L t)dt = \frac{b-a}{T} \int_0^Tf(u)du$

Question

Show if $f : \mathbb{R} \rightarrow \mathbb{R}$ is $T$-periodic and continuous then $$\lim\limits_{L \rightarrow +\infty} \int_a^b f(L t)dt = \frac{b-a}{T} \int_0^Tf(u)du$$ where : $a,b\in\mathbb{R}, a<b$, and $L$ is real.

EDIT : What I've done :

Let $k = \max \{k\in\mathbb{Z},La + kT \leq Lb\} = k = \lfloor{\frac{L(b-a)}{T}}\rfloor$. Then : $ \int_{La}^{La + kT} f(u)du = \int_0^{kT}f(u)du$ . I have to show $$\lim\limits_{L \rightarrow +\infty} \frac{1}{L}\int_{La + kT}^{Lb} f(u)du = 0$$ I have some difficulties to show that part.