Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

We have, $\underset{x\to 0}{lim}\frac{2^x-1}{(1+x{)}^{\frac{1}{2}}-1}$

At x=0, the value of the expression is the form of $\frac{0}{0}$.

Applying L'Hospital Rule

$\underset{x\to 0}{lim}\frac{\frac{d}{dx}(2^x-1)}{\frac{d}{dx}\left[\right(1+x{)}^{\frac{1}{2}}-1]}$

$\underset{x\to 0}{lim}\frac{2^{x}lo{g}_e^2-0}{\frac{1}{2}(1+x{)}^{\frac{1}{2}-1}.\frac{d}{dx}(1+x)-0}$

$\underset{x\to 0}{lim}\frac{2^{x}lo{g}_e^2}{\frac{1}{2\sqrt{(1+x)}}xL}$

$\underset{x\to 0}{lim}2(\sqrt{(1+x)}{2}^x.lo{g}_e^2$

Putting x=0

$=2\left(\sqrt{1+0}\right)2^{0}lo{g}_e^2$

$=2\times 1\times 1.lo{g}_e^2$

$=2lo{g}_e^2=lo{g}_e^4$