题目：求极限 \(\lim _{x\rightarrow0}\frac{1-\prod_{i=1}^{n}\cos(ix)}{x^2}\)
解答：
\[\begin{align} 对\cos(ix)在\ x=0\ 附近进行二阶泰勒展开，得\cos(ix)=1-\frac12x^2+o(x^2) \\则有\prod_{i=1}^{n}\cos(ix)=\prod_{i=1}^{n}[1-\frac12i^2x^2+o(i^2x^2)]=1-\frac12\sum_{i=1}^{n}i^2x^2+o(x^2) \\\Longrightarrow原式=\frac12\lim_{x\rightarrow0} \frac{(\sum_{i=1}^{n}i^2x^2)+o(x^2)}{x^2}=\frac12\sum_{i=1}^{n}i^2=\frac12 \frac{n(n+1)(2n+1)}{6} \\=\frac{n(n+1)(2n+1)}{12} \end{align}\]