I need help with the following question:
Let $f: [- \pi, \pi] \rightarrow \mathbb{R}$ be a differentiable function such that $f(0)=0$. Show that $\lim\limits_{N \to \infty} \int^{\pi}_{-\pi} f(t) \frac{\sin \left( \left( N+\frac{1}{3}\right)t \right)}{t} dt = 0$
Here's what I've tried: using L'Hôpital's, we can see that for each $N \in \mathbb{N}$, $\lim\limits_{t \to 0} f(t) \frac{\sin \left( \left( N+\frac{1}{3}\right)t \right)}{t} = 0 \cdot \left( N+\frac{1}{3} \right) = 0$. Then $\forall \varepsilon >0 ~ \exists \delta >0. ~ x \in (-\delta, \delta) \rightarrow f(x) \in (-\varepsilon, \varepsilon)$. Then I wanted to show that $\lim\limits_{N \to \infty} \int^{\pi}_{\delta} f(t) \frac{\sin \left( \left( N+\frac{1}{3}\right)t \right)}{t} dt \leq \frac{1}{\delta} \int^{\pi}_{\delta} f(t) \sin \left( \left( N+\frac{1}{3}\right)t \right) dt \xrightarrow[]{N \to \infty} 0$, maybe using Riemann-Lebesgue. The problem is that $\delta$ is different for every $N$, and also I'm not sure Riemann-Lesbegue can be used like that. Any help would be appriciated, thanks in advance!