Question

Evaluate the limit:

$\underset{x\to 2}{lim}\frac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2}$

Answer 1

At $x\to 2$,

$\frac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2}\text{becomes}\frac{0}{0}form$

$\underset{x\to 2}{lim}\frac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2}$

On rationalising numerator, we get

$=\underset{x\to 2}{lim}\frac{(\sqrt{1+4x}-\sqrt{5+2x})(\sqrt{1+4x}+\sqrt{5+2x})}{(x-2)(\sqrt{1+4x}+\sqrt{5+2x})}$

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

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

$=\underset{x\to 2}{lim}\frac{(1+4x-5-2x)}{(x-2)(\sqrt{1+4x}+\sqrt{5+2x})}$

$=\underset{x\to 2}{lim}\frac{(2x-4)}{(x-2)(\sqrt{1+4x}+\sqrt{5+2x})}$

$=\underset{x\to 2}{lim}\frac{2(x-2)}{(x-2)(\sqrt{1+4x}+\sqrt{5+2x})}\left[\right(x-2)\ne 0]$

$=\underset{x\to 2}{lim}\frac{2}{(\sqrt{1+4x}+\sqrt{5+2x})}$

$=\frac{2}{(\sqrt{1+4\left(2\right)}+\sqrt{5+2\left(2\right)})}$

$\frac{2}{(\sqrt{9}+\sqrt{9})}$

$=\frac{2}{3+3}$

$=\frac{2}{6}$

$=\frac{1}{3}$

$Therefore,$

$\underset{x\to 2}{lim}\frac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2}=\frac{1}{3}$