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. None of these

Answer 1

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

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

${lim}_{x\to 2}\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\times \frac{\sqrt{1+\sqrt{2+x}}+\sqrt{3}}{\sqrt{1+\sqrt{2+x}}+\sqrt{3}}$

$=\underset{x\to 2}{lim}\frac{\sqrt{2+x}-2}{\left(\sqrt{(\sqrt{1}+\sqrt{2+x}+\sqrt{3})(x-2)}\right)}$$=\underset{x\to 2}{lim}\frac{(x-2)}{(\sqrt{1+\sqrt{2+x}}+\sqrt{3})(\sqrt{2+x}+2)(x-2)}\text{(Rationalizing)}$

$=\frac{1}{\left(2\sqrt{3}\right)4}=\frac{1}{8\sqrt{3}}$
Hence, option 'A' is correct.