Question

Evaluate $\underset{x\to 8}{lim}\frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$

• A. $\frac{3}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{24}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{24}$

$\underset{x\to 8}{lim}\frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$

$=\underset{x\to 8}{lim}\frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}\cdot \frac{\sqrt{1+\sqrt{1+x}}+2}{\sqrt{1+\sqrt{1+x}}+2}$

$=\underset{x\to 8}{lim}\frac{\sqrt{1+x}-3}{x-8}\cdot \frac{1}{\sqrt{1+\sqrt{1+x}}+2}$

$=\underset{x\to 8}{lim}\frac{\sqrt{1+x}-3}{x-8}\cdot \frac{\sqrt{1+x}+3}{\sqrt{1+x}+3}\cdot \frac{1}{\sqrt{1+\sqrt{1+x}}+2}$

$=\underset{x\to 8}{lim}\frac{x-8}{x-8}\cdot \frac{1}{\sqrt{1+x}+3}\cdot \frac{1}{\sqrt{1+\sqrt{1+x}}+2}$

$=\underset{x\to 8}{lim}\frac{1}{\sqrt{1+x}+3}\cdot \frac{1}{\sqrt{1+\sqrt{1+x}}+2}$

$=\frac{1}{6}\cdot \frac{1}{4}=\frac{1}{24}$