Calculate the limit of $(1+x2^x)/(1+x3^x)$ to the power $1/x^2$ when $x\to 0$ I have a problem with this: $\displaystyle \lim_{x \rightarrow 0}{\left(\frac{1+x2^x}{1+x3^x}\right)^\frac{1}{x^2}}$. 
I have tried to modify it like this: $\displaystyle\lim_{x\rightarrow 0}{e^{\frac{1}{x^2}\ln{\frac{1+x2^x}{1+x3^x}}}}$ and then calculate the limit of the exponent: $\displaystyle \lim_{x\rightarrow 0}{\frac{1}{x^2}\ln{\frac{1+x2^x}{1+x3^x}}}$. 
But I don't know what to do next. Any ideas?

As usual we take logs here because we have an expression of the form $\{f(x) \} ^{g(x)} $. We can proceed as follows
\begin{align}
\log L&=\log\left\{\lim_{x\to 0}\left(\frac{1+x2^{x}}{1+x3^{x}}\right)^{1/x^{2}}\right\}\notag\\
&=\lim_{x\to 0}\log\left(\frac{1+x2^{x}}{1+x3^{x}}\right)^{1/x^{2}}\text{ (via continuity of log)} \notag\\
&=\lim_{x\to 0}\frac{1}{x^{2}}\log\frac{1+x2^{x}}{1+x3^{x}}\notag\\
&=\lim_{x\to 0}\frac{1}{x^{2}}\cdot\dfrac{\log\dfrac{1+x2^{x}}{1+x3^{x}}}{\dfrac{1+x2^{x}}{1+x3^{x}}-1}\cdot\left(\frac{1+x2^{x}}{1+x3^{x}}-1\right)\notag\\
&=\lim_{x\to 0}\frac{1}{x^{2}}\cdot\frac{x(2^{x}-3^{x})}{1+x3^{x}}\notag\\
&=\lim_{x\to 0}\frac{2^{x}-3^{x}}{x}\notag\\
&=\lim_{x\to 0}\frac{2^{x}-1}{x}-\frac{3^{x}-1}{x}\notag\\
&=\log 2-\log 3\notag\\
&=\log(2/3)\notag
\end{align} 
and hence $L=2/3$. We have used the following standard limits here $$\lim_{x\to 1}\frac{\log x} {x-1}=1,\,\lim_{x\to 0}\frac{a^{x}-1}{x}=\log a$$ There is no need of more powerful tools like L'Hospital's Rule or Taylor's theorem. 
