<s>[INST] 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?
 [/INST] let $f(x)=\ln\left(\frac{1+x2^x}{1+x3^x}\right)$ and $g(x)=x^2$ then we have $0/0$ for $x$ tends to $0$ you must calculate $$\frac{f'(x)}{g'(x)}$$ and look if the limit exists.
you will get this here $$1/2\,{\frac {{2}^{x}\ln  \left( 2 \right) {3}^{x}{x}^{2}-{2}^{x}{3}^{x
}\ln  \left( 3 \right) {x}^{2}+x{2}^{x}\ln  \left( 2 \right) -x{3}^{x}
\ln  \left( 3 \right) +{2}^{x}-{3}^{x}}{ \left( 1+x{2}^{x} \right) 
 \left( 1+x{3}^{x} \right) x}}
$$
and now L'Hospital one more times
the next quotient is $${\frac {{2}^{x} \left( \ln  \left( 2 \right)  \right) ^{2}{3}^{x}{x}^{
2}+2\,{2}^{x}\ln  \left( 2 \right) {3}^{x}x-{2}^{x}{3}^{x} \left( \ln 
 \left( 3 \right)  \right) ^{2}{x}^{2}-2\,{2}^{x}{3}^{x}\ln  \left( 3
 \right) x+2\,{2}^{x}\ln  \left( 2 \right) +x{2}^{x} \left( \ln 
 \left( 2 \right)  \right) ^{2}-2\,{3}^{x}\ln  \left( 3 \right) -x{3}^
{x} \left( \ln  \left( 3 \right)  \right) ^{2}}{2\,{2}^{x}\ln  \left( 
2 \right) {3}^{x}{x}^{3}+2\,{2}^{x}{3}^{x}\ln  \left( 3 \right) {x}^{3
}+2\,{2}^{x}\ln  \left( 2 \right) {x}^{2}+6\,{2}^{x}{3}^{x}{x}^{2}+2\,
{3}^{x}\ln  \left( 3 \right) {x}^{2}+4\,x{2}^{x}+4\,x{3}^{x}+2}}
$$ and here you can set $x=0$ the result it $$\ln(2)-\ln(3)$$
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