Answer
$$
\lim _{x \rightarrow 0} \frac{x 3^{x}}{3^{x}-1} =\frac{1}{\ln 3}
$$
Work Step by Step
$$
\lim _{x \rightarrow 0} \frac{x 3^{x}}{3^{x}-1}
$$
Since
$$
\lim _{x \rightarrow 0}( x 3^{x})=0
$$
and
$$
\lim _{x \rightarrow 0}( 3^{x}-1)=0
$$
the limit is an indeterminate form of type $\frac{0}{0}$ so we can apply l’Hospital’s Rule:
$$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{x 3^{x}}{3^{x}-1} & \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 0} \frac{x 3^{x} \ln 3+3^{x}}{3^{x} \ln 3} \\
&=\lim _{x \rightarrow 0} \frac{3^{x}(x \ln 3+1)}{3^{x} \ln 3} \\
&=\lim _{x \rightarrow 0} \frac{x \ln 3+1}{\ln 3} \\
&=\frac{1}{\ln 3}
\end{aligned}
$$
