Question

Use the formula $\underset{x\to 0}{lim}\frac{a^x-1}{x}=\log a$ to find $\underset{x\to 0}{lim}\frac{2^x-1}{{(1+x)}^{1/2}-1}$ which is $=a\log a$. Find $a$

Answer 1

$\underset{x\to 0}{lim}\frac{2^x-1}{{(1+x)}^{1/2}-1}$
$=\underset{x\to 0}{lim}\frac{(2^x-1)\{{(1+x)}^{1/2}+1\}}{\{{(1+x)}^{1/2}-1\}\{{(1+x)}^{1/2}+1\}}$
$=\underset{x\to 0}{lim}\frac{2^x-1}{1+x-1}\{{(1+x)}^{1/2}+1\}$
$=\underset{x\to 0}{lim}\frac{2^x-1}{x}\underset{x\to 0}{lim}\{{(1+x)}^{1/2}+1\}$
$=\left(\log 2\right).2=2\log 2.$