Question

Evaluate the limit:

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

Answer 1

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

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

On rationalising numerator, we get

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

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

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

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

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

$=\underset{x\to 2}{lim}\frac{(x+2)(x-2)}{(x-2)(\sqrt{x^2+1}+\sqrt{5})}\left[\right(x-2)\ne 0]$

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

$=\frac{2+2}{\sqrt{2^2+1}+\sqrt{5}}$

$=\frac{4}{2\sqrt{5}}$


$=\frac{2}{\sqrt{5}}$