问题

Questions in category: 极限 (Limit).

证明 $\sqrt{x + 1} - \sqrt{x} = \frac{1}{2 \sqrt{x + θ (x)}}$

Posted by haifeng on 2015-01-11 19:36:08 last update 2015-01-11 19:36:08 | Answers (2) | 收藏

设 $x ⩾ 0$ , 证明

$\sqrt{x + 1} - \sqrt{x} = \frac{1}{2 \sqrt{x + θ (x)}},$

其中 $θ (x)$ 满足不等式 $\frac{1}{4} ⩽ θ (x) < \frac{1}{2}$ .