Question

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

(a) 1

(b) 0

(c) $\sqrt{a}$

(d) $\frac{1}{\sqrt{a}}$

Answer 1

$\text { } \begin{array}{l} \lim _{x \rightarrow 0}\left[\frac{\sqrt{a+x}-\sqrt{a-x}}{x}\right] \\=\lim _{x \rightarrow 0}\left[\frac{(\sqrt{a+x}-\sqrt{a-x}) \sqrt{a+x}+\sqrt{a-x})}{x(\sqrt{a+x}+\sqrt{a-x})}\right] \\=\lim _{x \rightarrow 0} \frac{(a+x)-(a-x)}{x(\sqrt{a+x}+\sqrt{a-x})} \\=\lim _{x \rightarrow 0}\left[\frac{2 x}{x(\sqrt{a+x}+\sqrt{a-x})}\right] \\=\frac{2}{\sqrt{a}+\sqrt{a}}=\frac{1}{\sqrt{a}}\end{array}$

So, The correct option of this question will be (D).