Question

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

Answer 1

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

Dividing the numerator and the denominator by x:

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{\log \left(1+x\right)}{x·\left({\displaystyle \frac{3^x-1}{x}}\right)}\right]\left[\because \underset{x\to 0}{\lim }\frac{\log \left(1+x\right)}{x}=1\underset{x\to 0}{\lim }\left(\frac{a^x-1}{x}\right)=\log a\right] \\ =\frac{1}{\log 3} \end{gathered}$$