Remarkable limit and L’Hôpital $\lim_{x\to0}\left(\frac{a^x-x\ln a}{b^x-x\ln b}\right)^{1/x^2}$ 
Compute
  $$
\lim_{x\to0}\left(\frac{a^x-x\ln a}{b^x-x\ln b}\right)^{1/x^2}
$$
  where $a$ and $b$ are positive numbers. 

I came to the two different forms of this limit, as $\lim_{x\to0}$

$$e^{\frac{\ln b-\ln b}{\ln a-\ln a+\ln b-\ln b}}$$
$$\frac{x^2\cdot a^x+1-\frac{a^2}x}{x^2\cdot b^x+1-\frac{b^2}x}$$
So, what I want to say is that I can't solve this problem and I'm here for any kind of help.
 A: Compute the limit of the logarithm of your function
$$
\lim_{x\to0}\ln\left(
  \left(\frac{a^x-x\ln a}{b^x-x\ln b}\right)^{1/x^2}
\right)=
\lim_{x\to0}
\frac{\ln(a^x-x\ln a)-\ln(b^x-x\ln b)}{x^2}
$$
under the assumption that $a\ne b$ (if $a=b$ the limit is obvious).
Here applying L’Hôpital is quite easy. If you find the limit is $l$, then the required limit is $e^l$.
Note that you can just compute
$$
l_a=\lim_{x\to0}\frac{\ln(a^x-x\ln a)}{x^2}
$$
because the given limit will then be $l_a-l_b$.
A: Assuming that the desired limit is $L$ we can proceed as follows:
\begin{align}
\log L &= \log\left\{\lim_{x \to 0}\left(\frac{a^{x} - x\log a}{b^{x} - x\log b}\right)^{1/x^{2}}\right\}\notag\\
&= \lim_{x \to 0}\log\left(\frac{a^{x} - x\log a}{b^{x} - x\log b}\right)^{1/x^{2}}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\log\left(1 + \frac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\frac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}\cdot\dfrac{\log\left(1 + \dfrac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}\right)}{\dfrac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}}\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\frac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}\notag\\
&= \lim_{x \to 0}\frac{a^{x} - b^{x} - x\log a + x\log b}{x^{2}}\tag{1}\\
&= \lim_{x \to 0}\frac{a^{x}\log a - b^{x}\log b - \log a + \log b}{2x}\text{ (via L'Hospital's Rule)}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\log a\cdot\frac{a^{x} - 1}{x} - \log b\cdot\frac{b^{x} - 1}{x}\notag\\
&= \frac{(\log a)^{2} - (\log b)^{2}}{2}\notag
\end{align}
If you prefer the easier technique of Taylor series then you need to apply it after step marked $(1)$ and use the expansion $$a^{x} = 1 + x\log a + \frac{x^{2}}{2}(\log a)^{2} + o(x^{2})$$ This way you get the answer immediately.
Thus finally $$L = \exp\left(\frac{(\log a)^{2} - (\log b)^{2}}{2}\right)$$
