<s>[INST] Easy way of Compute this limit Easy way to compute: $$\lim_{x\to \:0\:}\left(\left(\frac{a^x-x\cdot \ln\left(a\right)}{b^x-x\cdot \ln\left(b\right)}\right)^{\frac{1}{x^2}}\right)$$
 [/INST] Let the desired limit be $L$. Then we have
\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(\frac{a^{x} - x\log a}{b^{x} - x\log b}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\log\left(1 + \frac{a^{x} - x\log a}{b^{x} - x\log b} - 1\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\log\left(1 + \frac{a^{x} - b^{x} - x(\log a - \log b)}{b^{x} - x\log b}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\dfrac{a^{x} - b^{x} - x(\log a - \log b)}{b^{x} - x\log b}\cdot\dfrac{\log\left(1 + \dfrac{a^{x} - b^{x} - x(\log a - \log b)}{b^{x} - x\log b}\right)}{\dfrac{a^{x} - b^{x} - x(\log a - \log b)}{b^{x} - x\log b}}\notag\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\dfrac{a^{x} - b^{x} - x(\log a - \log b)}{b^{x} - x\log b}\cdot 1\notag\\
&= \lim_{x \to 0}\frac{a^{x} - b^{x} - x(\log a - \log b)}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{a^{x} - 1 - x\log a}{x^{2}} - \frac{b^{x} - 1 - x\log b}{x^{2}}\notag\\
&= \lim_{x \to 0}(\log a)^{2}\cdot\frac{a^{x} - 1 - x\log a}{(x\log a)^{2}} - (\log b)^{2}\cdot\frac{b^{x} - 1 - x\log b}{(x\log b)^{2}}\notag\\
&= (\log a)^{2}\lim_{u \to 0}\frac{e^{u} - 1 - u}{u^{2}} - (\log b)^{2}\lim_{v \to 0}\frac{e^{v} - 1 - v}{v^{2}}\text{ (putting }u = x\log a, v = x\log b)\notag\\
&= \{(\log a)^{2} - (\log b)^{2}\}\lim_{u \to 0}\frac{e^{u} - 1 - u}{u^{2}}\notag
\end{align}
The last limit is easily seen to be $1/2$ either via one application of L'Hospital's Rule or through Taylor's series. Hence $$\log L = \frac{(\log a)^{2} - (\log b)^{2}}{2}$$ and therefore $$L = \exp\left(\frac{(\log a)^{2} - (\log b)^{2}}{2}\right)$$
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