Show that $\lim_{k\to\infty} \int_0^1 f(x) \sin(kx) \, dx = 0$

Assume function $f(x)$ has maximum $M$ and minimum $m$ in interval $[0,1].$ Since the integral exists so both $m$ and $M$ are real and bounded otherwise we will have counterexamples such as $f(x)={1\over x^2}$. Now we want to show that:

$$\lim_{k\to\infty} \int_0^1{f(x)\sin kx} \, dx=0$$

We try to bound the above integral. Since $m\le f(x)\le M$:

$$m\int_0^1 \sin k x \, dx \le \int_0^1f(x)\sin kx \,dx \le M \int_0^1\sin k x \, dx$$ or $$m{1-\cos k \over k}\le \int_0^1 f(x)\sin k x \, dx \le M{1-\cos k \over k}$$ By tending $k\to\infty$ both bounds get zero and we obtain what we want.