用无穷级数解决一道极限题

题目如下：

\begin{array}{r} lim_{x \to 0} \frac{x - \ln (1 + x)}{\sqrt{1 + x \sin x} - \sqrt{\cos x}} \end{array}

$\frac{0}{0}$ 类型的题目，可以用等价无穷小替换来解决，这里使用无穷级数，可以很快的出结论：

\begin{aligned} \ln (1 + x) = x - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} - \frac{1}{4} x^{4} + \dots \\ \sin x = x - \frac{1}{3!} x^{3} + \frac{1}{5!} x^{5} - \frac{1}{7!} x^{7} + \dots \\ \cos x = 1 - \frac{1}{2!} x^{2} + \frac{1}{4!} x^{4} - \frac{1}{6!} x^{6} + \dots \end{aligned}

以上推理可见这里（三角函数无穷级数展开）和这里（对数函数无穷级数展开）。

原式分母有理化，代入展开的无穷级数得：

\begin{aligned} lim_{x \to 0} \frac{x - \ln (1 + x)}{\sqrt{1 + x \sin x} - \sqrt{\cos x}} & = lim_{x \to 0} \frac{x - \ln (1 + x)}{1 + x \sin x - \cos x} * (\sqrt{1 + x \sin x} + \sqrt{\cos x}) \\ = lim_{x \to 0} \frac{x - \ln (1 + x)}{1 + x \sin x - \cos x} * lim_{x \to 0} (\sqrt{1 + x \sin x} + \sqrt{\cos x}) \\ = 2 lim_{x \to 0} \frac{x - \ln (1 + x)}{1 + x \sin x - \cos x} \\ = 2 lim_{x \to 0} \frac{x - (x - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} - \frac{1}{4} x^{4} + \dots)}{1 + x (x - \frac{1}{3!} x^{3} + \frac{1}{5!} x^{5} - \frac{1}{7!} x^{7} + \dots) - (1 - \frac{1}{2!} x^{2} + \frac{1}{4!} x^{4} - \frac{1}{6!} x^{6} + \dots)} \\ = 2 lim_{x \to 0} \frac{\frac{1}{2} x^{2} - \frac{1}{3} x^{3} + \frac{1}{4} x^{4} - \dots}{x^{2} - \frac{1}{3!} x^{4} + \frac{1}{5!} x^{6} - \frac{1}{7!} x^{8} + \dots + \frac{1}{2!} x^{2} - \frac{1}{4!} x^{4} + \frac{1}{6!} x^{6} - \dots} \\ = 2 lim_{x \to 0} \frac{\frac{1}{2} - \frac{1}{3} x + \frac{1}{4} x^{2} - \dots}{1 - \frac{1}{3!} x^{2} + \frac{1}{5!} x^{4} - \frac{1}{7!} x^{6} + \dots + \frac{1}{2!} - \frac{1}{4!} x^{2} + \frac{1}{6!} x^{4} - \dots} \\ = 2 * \frac{\frac{1}{2}}{1 + \frac{1}{2}} \\ = \frac{2}{3} \end{aligned}

无穷级数的思想真是妙啊！