Question

Evaluate the limit:

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

Answer 1

At $x\to 2,\frac{\sqrt{3-x}-1}{2-x}\text{becomes}\frac{0}{0}form$

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

On rationalising numerator, we get

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

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

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

$=\underset{x\to 2}{lim}\frac{(2-x)}{(2-x)(\sqrt{3-x}+1)}\left[\right(2-x)\ne 0]$

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

$=\frac{1}{(\sqrt{3-2}+1)}=\frac{1}{2}$

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