Question

$\underset{x\to 3}{\lim }\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}$

Answer 1

$\underset{x\to 3}{\lim }\left[\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right]$

It is of form $\frac{0}{0}.$

Rationalising the denominator:

$\underset{x\to 3}{\lim }\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{\left(\sqrt{x-2}-\sqrt{4-x}\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}\right]$

= $\underset{x\to 3}{\lim }\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{\left(x-2\right)-\left(4-x\right)}\right]$

= $\underset{x\to 3}{\lim }\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{2x-6}\right]$

= $\underset{x\to 3}{\lim }\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{2\left(x-3\right)}\right]$

= $\frac{\sqrt{3-2}+\sqrt{4-3}}{2}$

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