<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] For any $a,b > 0$ and as $x \to 0$:
$\def\l{\!\left}$
$\def\r{\right}$
  $a^x = \exp(x\ln(a)) \in 1 + x\ln(a) + \frac{1}{2}(x\ln(a))^2 + o(x^2)$.
  $b^x = \exp(x\ln(b)) \in 1 + x\ln(b) + \frac{1}{2}(x\ln(b))^2 + o(x^2)$.
  $\l( \dfrac{a^x - x\ln(a)}{b^x - x\ln(b)} \r)^\dfrac{1}{x^2} = \exp\l( \dfrac{1}{x^2}\ln\l( \dfrac{a^x - x\ln(a)}{b^x - x\ln(b)} \r) \r)$
  $ \in \exp\l( \dfrac{1}{x^2}\ln\l( \dfrac{ 1 + \frac{1}{2}\ln(a)^2 x^2 + o(x^2) }{ 1 + \frac{1}{2}\ln(b)^2 x^2 + o(x^2) } \r) \r)$
  $ \subseteq \exp\l( \dfrac{1}{x^2}\ln\l( 1 + \frac{1}{2}(\ln(a)^2-\ln(b)^2) x^2 + o(x^2) \r) \r)$
  $ \subseteq \exp\l( \dfrac{1}{x^2} \l( \frac{1}{2}(\ln(a)^2-\ln(b)^2) x^2 + o(x^2) \r) \r)$
  $ \subseteq \exp\l( \frac{1}{2}(\ln(a)^2-\ln(b)^2) + o(1) \r) = \exp\l( \frac{1}{2}(\ln(a)^2-\ln(b)^2) \r) ( 1 + o(1) )$
  $ \to \exp\l( \frac{1}{2}(\ln(a)^2-\ln(b)^2) \r)$.
</s>