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}

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 $A\approx 0.852248$.
