Question

The value of $\underset{x\to 0}{lim}\left[\frac{\sqrt{a+x}-\sqrt{a-x}}{x}\right]$ is

• A. $1$
• B. $0$
• C. $\sqrt{a}$
• D. $\frac{1}{\sqrt{a}}$

Answer 1

Answer: (D)

$\frac{1}{\sqrt{a}}$

$\underset{x\to 0}{lim}\left[\frac{\sqrt{a+x}-\sqrt{a-x}}{x}\right]$
$=\underset{x\to 0}{lim}\left[\frac{(\sqrt{a+x}-\sqrt{a-x})(\sqrt{a+x}+\sqrt{a-x})}{x(\sqrt{a+x}+\sqrt{a-x})}\right]$
$=\underset{x\to 0}{lim}\left[\frac{(a+x)-(a-x)}{x(\sqrt{a+x}+\sqrt{a-x})}\right]$
$=\underset{x\to 0}{lim}\left[\frac{2x}{x(\sqrt{a+x}+\sqrt{a-x})}\right]$
$=\underset{x\to 0}{lim}\left[\frac{2}{(\sqrt{a+x}+\sqrt{a-x})}\right]$
$=\frac{2}{\sqrt{a}+\sqrt{a}}=\frac{1}{\sqrt{a}}$