原理区
先看例子：
\(\lim_{{x \to 0}} \frac{{1 - \sqrt{\cos x}}}{{x^2}}\)
\(=-\lim_{{x \to 0}} \frac{{ln(1 + (\sqrt{\cos x} - 1)}}{{x^2}}\) //逆向代等价无穷小
\(=-\lim_{{x \to 0}} \frac{{ln(\sqrt{\cos x)}}}{{x^2}}\)
\(=-\frac{1}{2}\lim_{{x \to 0}} \frac{{ln(1+(cos x - 1))}}{{x^2}}\) //等价无穷小
\(=-\frac{1}{2}\lim_{{x \to 0}} \frac{{-\frac{1}{2}x^2}}{{x^2}}\)
\(=\frac{1}{4}\)
由此拓展出求该极限：
\(\lim_{{x \to 0}} \frac{{1 - (cos x)^a}}{{x^2}}\)
\(=-\lim_{{x \to 0}} \frac{{ln(1+(cosx)^a-1)}}{{x^2}}\)
\(=-\lim_{{x \to 0}} \frac{{ln(cos)^a}}{{x^2}}\)
\(=-a\lim_{{x \to 0}} \frac{{ln(1+(cosx-1))}}{{x^2}}\)
\(=-a\lim_{{x \to 0}} \frac{{cosx-1}}{{x^2}}\)
\(=\frac{a}{2}\)
与上面例子过程类似，即可推广出：
\(\lim_{{x \to 0}}\) 时，\(1-(cosx)^a\)~\(\frac{a}{2}x^2\)
实战区(来源：武忠祥每日一题-39)
\(\lim_{{x \to 0}} \frac{[1-(cosx)^\frac{1}{2}][1-(cosx)^\frac{1}{3}]...[1-(cosx)^\frac{1}{n}]}{(\frac{1}{2}x^2)^(n-1)}\)
\(=\lim_{{x \to 0}} \frac{(\frac{1}{4}x^2)(\frac{1}{6}x^2)...(\frac{1}{2n}x^2)}{{(\frac{1}{2}x^2)^(n-1)}}\)
\(=\lim_{{x \to 0}} \frac{(\frac{1}{2})^{n-1}(\frac{1}{2}\frac{1}{3}...\frac{1}{n})x^{2(n-1)}}{(\frac{1}{2})^{n-1}x^{2(n-1)}}\)
\(=\frac{1}{n!}\)