Question

Evaluate the limit:

$\underset{x\to 1}{lim}\frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1}$

Answer 1

At $x\to 1$,

$\frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1}\text{becomes}\frac{0}{0}form$

$\underset{x\to 1}{lim}\frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1}$

On rationalising numerator, we get

$=\underset{x\to 1}{lim}\frac{(\sqrt{3+x}-\sqrt{5-x})(\sqrt{3+x}+\sqrt{5-x})}{(x^2-1)(\sqrt{3+x}+\sqrt{5-x})}$

$=\underset{x\to 1}{lim}\frac{({\left(\sqrt{3+x}\right)}^2-{\left(\sqrt{5-x}\right)}^2)}{(x^2-1)(\sqrt{3+x}+\sqrt{5-x})}$

$=\underset{x\to 1}{lim}\frac{\left(\right(3+x)-(5-x\left)\right)}{(x^2-1)(\sqrt{3+x}+\sqrt{5-x})}$

$=\underset{x\to 1}{lim}\frac{(3+x-5+x)}{(x^2-1)(\sqrt{3+x}+\sqrt{5-x})}$

$=\underset{x\to 1}{lim}\frac{2(x-1)}{(x-1)(x+1)(\sqrt{3+x}+\sqrt{5-x})}$

$\left[\right(x-1)\ne 0]$

$=\underset{x\to 1}{lim}\frac{2}{(x+1)(\sqrt{3+x}+\sqrt{5-x})}$

$=\frac{2}{(1+1)(\sqrt{3+1}+\sqrt{5-1})}$

$=\frac{2}{2(\sqrt{4}+\sqrt{4})}$

$=\frac{1}{2+2}$

$=\frac{1}{4}$

$Therefore,$

$\underset{x\to 1}{lim}\frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1}=\frac{1}{4}$