Question

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

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

Answer 1

The correct option is A

$\frac{1}{8\sqrt{3}}$

$\because {lim}_{x\to +2}\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}$
$={lim}_{x\to 2}\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\times \frac{\sqrt{1+\sqrt{2+x}}+\sqrt{3}}{\sqrt{1+\sqrt{2+x}}+\sqrt{3}}$
$={lim}_{x\to 2}\frac{\sqrt{2+x}-2}{(x-2)\left(\sqrt{1+\sqrt{2+x}+\sqrt{3}}\right)}$
$={lim}_{x\to 2}\frac{\sqrt{2+x}-2}{(x-2)\left(\sqrt{1+\sqrt{2+x}+\sqrt{3}}\right)}\times \frac{\sqrt{2+x}+2}{\sqrt{2+x}+2}$
$=\frac{1}{\left(\sqrt{1+\sqrt{2+x}+\sqrt{3}}\right)(\sqrt{2+2}+2)}$
$=\frac{1}{2\sqrt{3}\times 4}$
$=\frac{1}{8\sqrt{3}}$