Question

$\underset{x\to 0}{lim}\frac{\sqrt{1+x}-1}{x}$ equals to

• A. $1$
• B. $\frac{1}{2}$
• C. $0$
• D. none of these

Answer 1

The correct option is B $\frac{1}{2}$

$\underset{x\to 0}{lim}\frac{\sqrt{1+x}-1}{x}$

$=\underset{x\to 0}{lim}\frac{\sqrt{1+x}-1}{x}.\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$

$=\underset{x\to 0}{lim}\frac{\sqrt{1+x}-1}{x}.\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$

$=\underset{x\to 0}{lim}\frac{1+x-1}{x(\sqrt{1+x}+1)}$

$=\underset{x\to 0}{lim}\frac{1}{\sqrt{1+x}+1}=\frac{1}{2}$

Hence, option 'B' is correct.