Question

The value of $=\underset{x\to 2}{\lim }\frac{\left[\sqrt{1+\sqrt{2+x}}-\sqrt{3}\right]}{x-2}$ is

• A. $\frac{1}{8\sqrt{3}}$
• B. $\frac{1}{4\sqrt{3}}$
• C. $0$
• D. None of these

Answer 1

Answer: (A)

The required limit $=\underset{x\to 2}{\lim }\frac{\left[1+\sqrt{2+x}-3\right]}{\left(x-2\right)\left[\sqrt{1+\sqrt{2+x}+\sqrt{3}}\right]}$
$=\underset{x\to 2}{\lim }\frac{\left(\sqrt{x+2}-2\right)\left(\sqrt{x+2}+2\right)}{\left(x-2\right)\left(\sqrt{1+\sqrt{2+x}+\sqrt{3}}\right)\left(\sqrt{x+2}+2\right)}$
$=\underset{x\to 2}{\lim }\frac{\left(x+2\right)-4}{\left(x-2\right)\left(\sqrt{1+\sqrt{2+x}+\sqrt{3}}\right)\left(\sqrt{x+2}+2\right)}$
$=\underset{x\to 2}{\lim }\frac{1}{\left(\sqrt{1+\sqrt{2+x}+\sqrt{3}}\right)\left(\sqrt{x+2}+2\right)}$
$=\frac{1}{2\sqrt{3}}.\frac{1}{4}=\frac{1}{8\sqrt{3}}$