Question

Evaluate the limit:

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

Answer 1

$Atx\to 0,\frac{\sqrt{1+x}-\sqrt{1-x}}{2x}\text{becomes}\frac{0}{0}form$

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

On rotaionalising numerator, we get

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

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

$[\because (a-b\left)\right(a+b)=(a^2-b^2\left)\right]$

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

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

$=\underset{x\to 0}{lim}\frac{1}{(\sqrt{1+x}+\sqrt{1-x})}$

$=\frac{1}{(\sqrt{1+0}+\sqrt{1-0})}$

$=\frac{1}{2}$

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