Question

$\underset{x\to 0}{lim}\frac{2^x-1}{{(1+x)}^{1/2}-1}$ is equal to

• A. $\log 2$
• B. $\log 4$
• C. $\log \sqrt{2}$
• D. None of these

Answer 1

The correct option is B $\log 4$

We need to find limit of $\underset{x\to 0}{lim}\frac{2^x-1}{{(1+x)}^{1/2}-1}$

Using L'hospital's rule, we get

$\underset{x\to 0}{lim}\frac{2^x\log 2}{\frac{1}{2}{(1+x)}^{-1/2}}$
$=2\log 2[\because \underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{lim}\frac{f^{{}^{\prime }}\left(x\right)}{g^{{}^{\prime }}\left(x\right)}]$
$=\log 4$