Question

Find the limit:-
$\underset{x\to 3}{lim}\left[\frac{\sqrt{x^2+1}-\sqrt{10}}{x-3}\right]$

Answer 1

$\underset{x\to 3}{lim}\left[\frac{\sqrt{x^2+1}-\sqrt{10}}{x-3}\right]$ by rationalising the numerator

$\underset{x\to 3}{lim}\frac{[\sqrt{x^2+1}-\sqrt{10}}{(x-3)}\times \frac{(\sqrt{x^2+1}+\sqrt{10})}{(\sqrt{x^2+1}+\sqrt{10})}$

$\underset{x\to 3}{lim}\frac{x^2+1-10}{(x-3(\sqrt{x^2+1}+\sqrt{10})}$

$\underset{x\to 3}{lim}\frac{x^2-9}{(x-3)(\sqrt{x^2+1}+\sqrt{10}}$

$\underset{x\to 3}{lim}\frac{(x-3)(x+3)}{(x-3)(\sqrt{x^2+1}+\sqrt{10}}$

$\frac{3+3}{\sqrt{3^2+1}+\sqrt{10}}=\frac{6}{\sqrt{10}+\sqrt{10}}=\frac{6}{2\sqrt{10}}=\frac{3}{\sqrt{10}}$