Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{x}{\sqrt{1+x}-\sqrt{1-x}}$

Answer 1

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

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

On rationalising denominator, we get

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

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

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

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

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

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

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

$=1$

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