Evaluating $\lim_{x\to0}\left(\frac{1+x2^x}{1+x3^x}\right)^\frac1{x^2}$ Find the following limit
$$
\lim_{x\to0}\left(\frac{1+x2^x}{1+x3^x}\right)^\frac1{x^2}
$$
I have used natural logarithm to get
$$
\exp\lim_{x\to0}\frac1{x^2}\ln\left(\frac{1+x2^x}{1+x3^x}\right)
$$
After this, I have tried l'opital's rule but I was unable to get it to a simplified form.
How should I proceed from here? Any here is much appreciated!

Solution Procedure Using Taylor Series and L'Hoptial
Try to understand or prove each of the following steps:
1) $\ln \left( {\frac{{1 + x{2^x}}}{{1 + x{3^x}}}} \right) = \ln \left( {1 + x\frac{{{2^x} - {3^x}}}{{1 + x{3^x}}}} \right)$  
2) $\ln (1 + x) = x + o({x^2})$
3) $o\left( {{x^2}{{\left( {{{{2^x} - {3^x}} \over {1 + x{3^x}}}} \right)}^2}} \right) = o\left( {o\left( {{x^2}} \right)} \right) = o({x^2})$
4) $\ln \left( {1 + x{{{2^x} - {3^x}} \over {1 + x{3^x}}}} \right) = x{{{2^x} - {3^x}} \over {1 + x{3^x}}} + o({x^2})$
5) $\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\ln \left( {1 + x{{{2^x} - {3^x}} \over {1 + x{3^x}}}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{1}{x^2} \times x{{{2^x} - {3^x}} \over {\left( {1 + x{3^x}} \right)}}$
6) $\mathop {\lim }\limits_{x \to 0} {{{2^x} - {3^x}} \over {x\left( {1 + x{3^x}} \right)}} = \mathop {\lim }\limits_{x \to 0} {{{2^x}\ln 2 - {3^x}\ln 3} \over {\left( {1 + x{3^x}} \right) + x\left( {1 + x\ln 3} \right){3^x}}} = \ln 2 - \ln 3 = \ln \left( {{2 \over 3}} \right)$
7) $ {\exp\lim_{x\to0}\frac1{x^2}\ln\left(\frac{1+x2^x}{1+x3^x}\right)} = {\exp(\ln{2 \over 3}})= {2 \over 3}$
