Question

Evaluate the limit:

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

Answer 1

At $x\to 0$,

$\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x}becomes\frac{0}{0}form$

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

On rationalising numerator, we get

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

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

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

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

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

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

$=0$