Question

Prove that
$\underset{x\to 0}{lim}\frac{\log \left(1+x\right)}{3^x-1}=$$\frac{1}{\log x}$

Answer 1

$\underset{x\to 0}{lt}\frac{log(1+x)}{3^x-1}$

$\Rightarrow \frac{0}{0}$ from (L hospital)$

$=\underset{x\to 0}{lt}\frac{\frac{d}{dx}\left(log\right(1+x\left)\right)}{\frac{d}{dx}(3^x-1)}$

$\Rightarrow \underset{x\to 0}{lt}\frac{\frac{1}{1+x}}{3^{x}logx}$

$\Rightarrow \frac{1}{3^{o}logx}$

$\Rightarrow \frac{1}{logx}$.