How to evaluate $\displaystyle\lim_{x\to0^+}\left(\frac{\ln(4^x-3^x)-\ln(4^x-1)}{x}\right)(4^x-1)$? How to evaluate
$$L:=\lim_{x\to0^+}\left(\frac{\ln(4^x-3^x)-\ln(4^x-1)}{x}\right)(4^x-1)$$?
My solution:
$$\begin{align}
L&=\lim_{x \to 0^+} \frac{\ln\left(\frac{4^x-3^x}{4^x-1}\right)}{x}(4^x-1)=\\
&=\lim_{x \to 0^+} \frac{ \ln\left(1-\frac{3^x-1}{4^x-1}\right)}{\frac{3^x-1}{4^x-1}}\times  \frac{3^x-1}{4^x-1}\times  \frac{4^x-1}{x}
\end{align}$$
How to continue from here?

$$\begin{align}L&=\lim_{x \to 0^+} \ln\left(1 - \frac{3^x-1}{4^x-1}\right)\times\frac{4^x-1}{x}=\\
&=\lim_{x \to 0^+} \ln\left(1 - \frac{3^x-1}{4^x-1}\right)\times\lim_{x \to 0^+}\frac{4^x-1}{x}=\\
&=\ln 4 \cdot \lim_{x \to 0^+} \ln\left(1 - \frac{3^x-1}{4^x-1}\right)=\\
&=\ln 4 \cdot \ln \left( \lim_{x \to 0^+}\left[1 - \frac{3^x-1}{4^x-1}\right]\right )=\tag{$\star$}\\
&=\ln 4 \cdot \ln \left(1 - \frac{\ln 3}{\ln 4} \right )\approx-2.17999444
\end{align}$$
In $(\star)$ we used the fact the logarithm is continuous in $1 - \frac{\ln 3}{\ln 4}$, so that we can evaluate the inner limit and then just substitute (there is a theorem about that, if you're curious).
Here is the plot of the function near $0$:

