求极限
\[ \lim_{n\to\infty}\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin{nx}}{\sin{x}}\right)^{4}dx \]
解:
首先注意到极限
\[ \lim_{x\to 0}x^2\left(\frac{1}{\sin^{4}{x}}-\frac{1}{x^{4}}\right)=\frac{2}{3} \]
则有
\[ \frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin{nx}}{\sin{x}}\right)^{4}dx=\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}(\sin{nx})^{4}\cdot\frac{1}{x}\cdot x^2\left(\frac{1}{\sin^{4}{x}}-\frac{1}{x^{4}}\right)dx+\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}\frac{(\sin{nx})^{4}}{x^{3}}dx=A_{1}+A_{2} \]
由于$F(x)=x^2\left(\frac{1}{\sin^{4}{x}}-\frac{1}{x^{4}}\right)$可以通过补充定义让它连续，则在$\left[0,\frac{\pi}{2}\right]$上$F(x)$有最大值$M$,而$(\sin{nx})^{4}\leq |\sin{nx}|\leq nx $,则
\[ A_{1}\leq \frac{M}{n}\to 0\qquad (n\to+\infty) \]
而
\begin{align*}
 \lim_{n\to\infty}\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}\frac{\sin^{4}nx}{x^{3}}dx&=\lim_{n\to+\infty}\int_{0}^{\frac{n\pi}{2}}\frac{\sin^{4}{x}}{x^{3}}dx\\
&=\int_{0}^{+\infty}\frac{\sin^{4}{x}}{x^{3}}dx\\
&=\int_{0}^{+\infty}\frac{2\sin^{3}{x}\cos{x}}{x^{2}}dx\\
&=\int_{0}^{+\infty}\frac{6\sin^{2}x\cos^2{x}-2\sin^{4}{x}}{x}dx\\
&=\int_{0}^{+\infty}\frac{\cos{2x}-\cos{4x}}{x}dx\\
&=\ln{2}
\end{align*}
因此
\[ \lim_{n\to\infty}\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin{nx}}{\sin{x}}\right)^{4}dx=\ln{2} \]