Question

Use the formula $\underset{x\to 0}{lim}\frac{a^x-1}{x}={\log }_{e}a$ to compute $\underset{x\to 0}{lim}\frac{2^x-1}{\sqrt{1+x}-1}$

Answer 1

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

$=\underset{x\to 0}{lim}\left[\begin{array}{c}\frac{2^x-1}{\sqrt{1+x}-1}\times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}\end{array}\right]$

$=\underset{x\to 0}{lim}\left(\begin{array}{c}\frac{2^x-1}{x}\end{array}\right)\times \underset{x\to 0}{lim}(\sqrt{1+x}+1)$

$={\log }_{e}2\times 2={\log }_{e}4$