Question

Find the following limit:
$\underset{x\to 4}{lim}\frac{\sqrt{1+2x}-3}{\sqrt{x}-2}.$

Answer 1

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

Using Rationalization method

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

$=\underset{x\to 4}{lim}\frac{(\sqrt{1+2x}-3)(\sqrt{x}+2)}{x-4}\times \frac{(\sqrt{1+2x}+3)}{(\sqrt{1+2x}+3)}$

$=\underset{x\to 4}{lim}\frac{(1+2x-9)(\sqrt{x}+2)}{(x-4)(\sqrt{1+2x}+3)}$

$=\underset{x\to 4}{lim}\frac{(2x-8)\left(\sqrt{x+2}\right)}{(x-4)(\sqrt{1+2}+3)}$

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

$=\frac{2(\sqrt{4}+2)}{(\sqrt{9}+3)}=\frac{2\times 4}{6}=\frac{4}{3}$.