Question

$\underset{x\to 0}{\lim }\frac{\sqrt{1+x}-\sqrt{1-x}}{2x}$

Answer 1

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

It is of the form $\frac{0}{0}.$

Rationalising the numerator:

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

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

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

= $\frac{1}{\sqrt{1+0}+\sqrt{1-0}}$
= $\frac{1}{2}$