Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$ 
Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$

Here is what I have done so far:
\begin{align}
\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}
&=
\lim_{x\to 0} \left(\frac{3^x-1}{4^x+2^x-2}+\frac{2^x-1}{4^x+2^x-2}\right)\\
&=\lim_{x\to 0} \left(\frac{3^x-1}{x}\frac{x}{4^x+2^x-2}+\frac{2^x-1}{x}\frac{x}{4^x+2^x-2}\right)\\
&=\ln 3\lim_{x\to 0} \frac{x}{4^x+2^x-2}+\ln 2\lim_{x\to 0} \frac{x}{4^x+2^x-2}
\end{align}
 A: Note that by standard limits since
$$\frac{a^x-1}{x}\to \log a$$
we have that
$$\frac{3^x+2^x-2}{4^x+2^x-2}=\frac{\frac{3^x-1}x+\frac{2^x-1}x}{\frac{4^x-1}x+\frac{2^x-1}x}\to\frac{\log 3+\log2}{\log 4+\log 2}=\frac{\log 6}{\log 8}$$
A: $$\lim_{x\rightarrow0}\frac{3^x+2^x-2}{4^x+2^x-2}=\lim_{x\rightarrow0}\frac{\frac{3^x-1}{x}+\frac{2^x-1}{x}}{\frac{4^x-1}{x}+\frac{2^x-1}{x}}=\frac{\ln3+\ln2}{\ln4+\ln2}=\log_86$$
I used $$\lim_{x\rightarrow0}\frac{a^x-1}{x}=\lim_{x\rightarrow0}\left(\frac{e^{x\ln{a}}-1}{x\ln{a}}\cdot\ln{a}\right)=\ln{a}$$ for all $a>0$.
A: \begin{align}\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}&=\lim_{x\to 0} \left(\frac{3^x-1}{4^x+2^x-2}+\frac{2^x-1}{4^x+2^x-2}\right)
\\&=\lim_{x\to 0} \left(\frac{3^x-1}{x}\frac{x}{4^x+2^x-2}+\frac{2^x-1}{x}\frac{x}{4^x+2^x-2}\right)
\\&=\ln 3\lim_{x\to 0} \frac{x}{4^x+2^x-2}+\ln 2\lim_{x\to 0} \frac{x}{4^x+2^x-2} \\&= (\ln 3 + \ln 2) \lim_{x\to 0} \frac{x}{4^x+2^x-2} 
\\&= (\ln 3 + \ln 2) \lim_{x\to 0} \frac{1}{\frac{4^x-1}{x}+\frac{2^x-1}{x}} \\
&=\frac{\ln 3 + \ln 2}{\ln 4 + \ln 2} \\
&= \frac{\ln 6}{\ln 8}
\end{align}
A: Just added for your curiosity since you already received good answers.
Since $$a^x=e^{x \log(a)}$$ using Taylor series around $x=0$
$$a^x=1+ \log (a)x+\frac{1}{2} \log ^2(a) x^2+O\left(x^3\right)$$ This makes
$$A=\frac{3^x+2^x-2}{4^x+2^x-2} =\frac{ (\log (2)+\log (3))x+\frac{1}{2}  \left(\log ^2(2)+\log
   ^2(3)\right)x^2+O\left(x^3\right) } {(\log (2)+\log (4))x+\frac{1}{2}  \left(\log ^2(2)+\log
   ^2(4)\right)x^2+O\left(x^3\right)  }$$ Divide top and bottom by $x$ and make the long division to get, after simplifications,
$$A=\frac{\log (6)}{\log (8)}-\frac{ \log \left(\frac{4}{3}\right) \log
   (54)}{18 \log (2)}x+O\left(x^2\right)$$ which shows the limit and how it is approached.
But you can use it for approximation. For illustration, use $x=0.1$. This would give $A\approx 0.852457$ while the exact value would be $A\approx 0.852248$.
