Question

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

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

Answer 1

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

We have,
$\underset{x\to 2}{lim}\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}$
$=\underset{x\to 2}{lim}\frac{1+\sqrt{2+x}-3}{(x-2)\{\sqrt{1+\sqrt{2+x}}+\sqrt{3}\}}$
$=\underset{x\to 2}{lim}\frac{\sqrt{2+x}-2}{(x-2)\{\sqrt{1+\sqrt{2+x}}+\sqrt{3}\}}$
$=\underset{x\to 2}{lim}\frac{(x-2)}{(x-2)\{\sqrt{1+\sqrt{2+x}}+\sqrt{3}\}\{\sqrt{2+x}+2\}}$
$=\underset{x\to 2}{lim}\frac{1}{\{\sqrt{1+\sqrt{2+x}}+\sqrt{3}\}\{\sqrt{2+x}+2\}}=\frac{1}{2\sqrt{3}\times 4}=\frac{1}{8\sqrt{3}}$
Hence, option 'A' is correct.