Question

The value of $\underset{x\to 2}{lim}\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}$ is

• A. $\frac{1}{8\sqrt{3}}$
• B. $\frac{1}{4\sqrt{3}}$
• C. $0$
• D. $2$

Answer 1

The correct option is A $\frac{1}{8\sqrt{3}}$

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

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