Question

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

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

Answer 1

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

Now,

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

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

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

$=\frac{1}{2}.1$ [ Since $\underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1$]

$=\frac{1}{2}$.