Question

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

Answer 1

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

It is of the form $\frac{0}{0}$.
Rationalising the numerator:

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

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

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

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

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

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