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!

$\lim_{x\to0} f(x)\\
\text{where } f(x)
=\left(\frac{1+xa^x}{1+xb^x}\right)^\frac1{x^2}
$
Let
$g(x, a)
=x a^x
$.
For small $x$,
$g(x, a)
=xe^{x \ln a}
\approx x(1+x \ln a + x^2 \ln^2a/2 + O(x^3))
= x(1+x \ln a + O(x^2))
$
so
$\begin{array}\\
\frac{1+xa^x}{1+xb^x}
&\approx \frac{1+x(1+x \ln a + O(x^2))}{1+x(1+x \ln b + O(x^2))}\\
&\approx (1+x(1+x \ln a + O(x^2)))(1-(x(1+x \ln b + O(x^2)))+x^2+O(x^2))\\
&= (1+x+x^2 \ln a + O(x^3))(1-x-x^2 (\ln b-1) + O(x^3))\\
&=1+x^2(\ln a-1 - (\ln b -1))+O(x^3)\\
&=1+x^2(\ln(a/b))+O(x^3)\\
\end{array}
$
so,
since
$(1+ax)^{1/x}
\approx e^{a}
$
as
$x \to 0$,
$f(x)
\approx (1+x^2(\ln(a/b))+O(x^3))^{1/x^2}
\approx e^{\ln(a/b)}
=\frac{a}{b}
$.
