Question

Evaluate the limit:
$\underset{x\to 3}{lim}\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}$

Answer 1

$Atx\to 3,\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\text{becomes}\frac{0}{0}form$

On rationalising denominator, we get

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

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

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

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

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

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

$=\underset{x\to 3}{lim}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2(x-3)}\left[\right(x-3)\ne 0]$

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

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

$=1$

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