Question

${lim}_{x\to 0}\frac{2^x-1}{\sqrt{1+x}-1}$ equals

• A. $\log 2$
• B. $\log 4$
• C. $\log 3$
• D. None of these

Answer 1

The correct option is B $\log 4$

$\begin{array}{c}{lim}_{x\to 0}\frac{2^x-1}{\sqrt{1+x}-1}\\ ={lim}_{x\to 0}\frac{\left(2^x-1\right)}{x}\times \frac{x}{\left(\sqrt{1+x}-1\right)}\\ ={lim}_{x\to 0}\frac{\left(2^x-1\right)}{x}\times \frac{x\times \left(\sqrt{1+x}+1\right)}{x}\\ ={lim}_{x\to 0}\frac{\left(2^x-1\right)}{x}\left(\sqrt{1+x}+1\right)\\ =2In2\\ =\log 4\\ OptionBiscorrectanswer.\end{array}$