Sorry, my mathematical english vocabulary is not as vivid as I would like it to be, therefore my topic touches the main problem in searching limit for this function.
$$\lim_{x \to 0} {\left (\frac{1+x\cdot2^x}{1+x\cdot3^x}\right)}^\frac{1}{x^2}$$
I tried several approaches, although it all ended up to having division by $0$ because of $3^x$ or $2^x$.
*EDIT: Can it be solved without using l'Hôpital's rule?
On
We compute
$$\lim_{x\to 0} \left(\frac{1+xe^{ax}}{1+xe^{bx}}\right)^{1/x^2},\qquad\qquad (\ast)$$
where $a=\log 2$ and $b=\log 3$, though that is not important. Note that
$$\log \left(\left(\frac{1+xe^{ax}}{1+xe^{bx}}\right)^{1/x^2}\right)=\frac{1}{x^2}\left(\log(1+xe^{ax})-\log(1+xe^{bx})\right).$$
Let $|x|$ be close to $0$. We find series expansions of everything. For the logarithm, we have
$$\log(1+t)=t-\frac{t^2}{2}+O(t^3).$$
Thus
$$\log(1+xe^{ax})=xe^{ax}-\frac{x^2e^{2ax}}{2}+O(x^3).$$
The beginning of the power series for the exponentials gives
$$\log(1+xe^{ax})=x+ax^2 -\frac{x^2}{2}+O(x^3),$$
and similarly
$$\log(1+xe^{bx})=x+bx^2 -\frac{x^2}{2}+O(x^3).$$
Subtract, and divide by $x^2$. We get
$$\frac{1}{x^2}\left(\log(1+xe^{ax})-\log(1+xe^{bx})\right)= a-b +O(x).$$
Thus
$$\lim_{x\to 0}\frac{1}{x^2}\left(\log(1+xe^{ax})-\log(1+xe^{bx})\right)=a-b,$$
and therefore the limit $(\ast)$ is $e^{a-b}$, which is $\dfrac{2}{3}$.
Without using l'Hopital:
Let $t=1/x$, so we want $$\lim_{t\to\infty}\left({t+2^{1/t}\over t+3^{1/t}}\right)^{t^2}$$ Then write $${t+2^{1/t}\over t+3^{1/t}}=1-{1\over r}{\rm\ where\ }r={t+3^{1/t}\over3^{1/t}-2^{1/t} }$$ Then $$\lim_{t\to\infty}(1-r^{-1})^r=e^{-1}$$ so we're looking at $\lim_{t\to\infty}e^{-t^2/r}$. Now I find it convenient to go back to $x$; we want $$\lim_{x\to0}e^{-(3^x-2^x)/((1+x3^x)x)}$$ We have $1+x3^x\to1$, and $(3^x-2^x)/x\to\log(3/2)$ (see below), so the whole thing goes to $e^{-\log(3/2)}=2/3$.
Let $f(z)=a^z$. Then we know that $f'(0)=\log a$, but by the difference-quotient definition, $f'(0)=\lim_{x\to0}(a^x-1)/x$. This explains $(3^x-2^x)/x\to\log(3/2)$.