find the limit of the function:
lim {ln(1+x)/(3^x -1)}
x-> 0
$\displaystyle \lim_{x \to 0} \frac{ \ln (1+x)}{(3^x -1)}=\lim_{x \to 0}\left \{ \frac{ \ln (1+x)}{x} \cdot \frac{x}{(3^x -1)} \right \}$
$\displaystyle = \lim_{x \to 0} \frac{ \ln (1+x)}{x} \cdot \frac{1}{\lim_{x \to 0} \frac{(3^x -1)}{x}}=1.\frac{1}{ \log_{e}{3}}=\frac{1}{\ln {3}}= \log_{3}{e}$
This is useless without an explanation of what you think you are doing.
Also
$\displaystyle \lim_{x \to 0} \frac{ \ln (1+x)}{x}=1$
needs to be justified as does:
$\displaystyle \lim_{x \to 0} \frac{(3^x -1)}{x}=\ln(3)$
Hence you may as well just have suggested the OP goes straight to L'Hopital's rule to find the limit.
And another thing, we prefer that hints or partial solutions are given, not full solutions or just the final answer.
CB
these are identities which are taught us in 12th standard
$\displaystyle \lim_{x \to 0} \frac{ \ln (1+x)}{x}=\lim_{x \to 0} \left ( \frac {x-\frac{x^2}{2}+\frac {x^3}{3}- .... }{x} \right ) $
$\displaystyle =\lim_{x \to 0} \left (1- \frac {x}{2} + \frac {x^2}{3} - .... \right ) =1$
and
$\displaystyle \lim_{x \to 0} \frac{(a^x -1)}{x} = \lim_{x \to 0} \left ( \frac {1+x log_{e}{a}+ \frac {x^2}{2!} (log_{e}{a})^2+.... -1}{x} \right )$
$\displaystyle = \lim_{x \to 0} \left ( log_{e}{a}+ \frac {x}{2!} (log_{e}{a})^2+.... \right ) = log_{e}{a} $
O.K.. CaptainAnd another thing, we prefer that hints or partial solutions are given, not full solutions or just the final answer.