📝 题目
例 10 求 $\displaystyle{\mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sqrt{1 + x} - 1}{x}}$ .
💡 答案与解析
解 令 $y = \sqrt{1 + x}$ ,则 $x = {y}^{2} - 1$ ,我们有
$$ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sqrt{1 + x} - 1}{x} = \mathop{\lim }\limits_{{y \rightarrow 1}}\frac{y - 1}{{y}^{2} - 1} = \mathop{\lim }\limits_{{y \rightarrow 1}}\frac{1}{y + 1} = \frac{1}{2}. $$