Suppose function $f\in\mathcal{C}^1[0,\frac{\pi}{2}]$, and $f(0)=0$, calculate:$$\lim\limits_{n\to\infty}\frac{1}{\ln n}\int_{0}^{\frac{\pi}{2}}\frac{\sin ^2nx}{\sin^2 x}f(x)\text d x$$
Because $f$ is continuous at $0$，there exists $\delta >0$ such that $\forall x\in (0,\delta)(|f(x)|<\varepsilon)$.
On the one hand, $\frac{\sin ^2nx}{\sin^2 x}f(x)$ is bounded when $x$ belongs to $[\delta,\frac{\pi}{2}]$，hence $$\lim\limits_{n\to\infty}\frac{1}{\ln n}\int_{\delta}^{\frac{\pi}{2}}\frac{\sin ^2nx}{\sin^2 x}f(x)\text d x=0$$
For the other side, we have $$\mid\frac{1}{\ln n}\int_{0}^{\delta}\frac{\sin ^2nx}{\sin^2 x}f(x)-f(0)\text d x\mid<\frac{\varepsilon}{\ln n}\int_{0}^{\delta}\frac{\sin ^2nx}{\sin^2 x}\text d x$$
But I don't know how to calculate $\int_{0}^{\delta}\frac{\sin ^2nx}{\sin^2 x}\text d x$, I have tried some methods but they don't work.