Answer
$$\lim_{x\to0}\frac{x3^x}{3^x-1}=\frac{1}{\ln3}$$
Work Step by Step
$$A=\lim_{x\to0}\frac{x3^x}{3^x-1}$$
For $\lim_{x\to0}(x3^x)=0\times3^0=0$ and $\lim_{x\to0}(3^x-1)=3^0-1=1-1=0,$
this is an indeterminate form of $\frac{0}{0}$. Applying L'Hospital's Rule, we have
$$A=\lim_{x\to0}\frac{(x3^x)'}{(3^x-1)'}$$
$$A=\lim_{x\to0}\frac{1\times3^x+x(3^x\ln3)}{3^x\ln3-0}$$
$$A=\lim_{x\to0}\frac{3^x+x3^x\ln3}{3^x\ln3}$$
$$A=\lim_{x\to0}\frac{3^x(1+x\ln3)}{3^x\ln3}$$
$$A=\lim_{x\to0}\frac{1+x\ln3}{\ln3}$$
Now we replace $x$ with $0$.
$$A=\frac{1+0\times\ln3}{\ln3}$$
$$A=\frac{1}{\ln3}$$
