Q. $\lim_{x \rightarrow 2}{= \frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2}}$ is equal to

A) $\frac{1}{8\sqrt{3}}$
B) $\frac{1}{\sqrt{3}}$
C) $8\sqrt{3}$
D) $\sqrt{3}$
Explanation

(a)
$\lim_{x \rightarrow 2}{= \frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2}}$
$= \lim_{x \rightarrow 2}\frac{(1 + \sqrt{2 + x} - 3)}{(x - 2)(\sqrt{1 + \sqrt{2 + x}} + \sqrt{3})}$
$= \lim_{x \rightarrow 2}\ \frac{x - 2}{(x - 2)\left( \sqrt{1 + \sqrt{2 + x}} + \sqrt{3} \right)(\sqrt{2 + x} + 2)}$
$= \lim_{x \rightarrow 2}\frac{1}{(\sqrt{1 + \sqrt{2 + x}} + \sqrt{3})(\sqrt{2 + x} + 2)}$
$= \frac{1}{\left( \sqrt{1 + 2} + \sqrt{3} \right)(\sqrt{2 + 2} + 2)} = \frac{1}{8\sqrt{3}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.