Easy way of Compute this limit Easy way to compute: $$\lim_{x\to \:0\:}\left(\left(\frac{a^x-x\cdot \ln\left(a\right)}{b^x-x\cdot \ln\left(b\right)}\right)^{\frac{1}{x^2}}\right)$$
 A: Apply $\ln$ to get 
$$\tag 1 \frac{\ln (a^x-x\ln a) - \ln (b^x-x\ln b)}{x^2}.$$
Using $e^u = 1 + u + u^2/2 +O(u^3)$ as $u\to 0,$ we get
$$a^x = e^{x\ln a} = 1 + x\ln a + (x\ln a)^2/2 + O(x^3).$$
A similar result holds for $b^x.$ Therefore $(1)$ equals
$$\tag 2 \frac{\ln (1+(x\ln a)^2/2 + O(x^3)) - \ln (1+(x\ln b)^2/2 + O(x^3))}{x^2}.$$
Now $\ln (1+u) = u + O(u^2).$ Therefore $(2)$ equals
$$\frac{(x\ln a)^2/2 -(x\ln b)^2/2 + O(x^3)}{x^2} \to (\ln a)^2/2 -(\ln b)^2/2 .$$
Exponentiating back gives $e^{(\ln a)^2/2 -(\ln b)^2/2}.$
A: Basically as $x \to 0$  the function $$\left(\frac{a^x -x \, \ln a}{b^x -x \, \ln b}\right)^{1/x^2}$$ attains the form 1^infinity. As $\lim x \to a  [f(x)]^{g(x)}$ attains $1^{\infty}$ form we can modify limit as $e^{\lim x \to a (f(x)-1) g(x)}$  
using the above rule, and after taking LCM your limit transforms to $e^{\lim x \to 0 (1/x^2){(a^x -x\ln a -b^x +x\ln b)/(b^x -x\ln b)}}$
  using expansion of $$a^x = 1 + (x\ln a)/1! + ((x\ln a)^2)/2! +.......$$
                     $$b^x = 1 +(x\ln b)/1!  + ((x\ln b)^2)/2! +....$$
your limit transforms to $$e^{\lim x-->0 (1/x^2){[((x\ln a)^2)/2!..... -((x\ln b)^2)/2!....]/(b^x -x\ln b)}}$$
cancel $x^2$ and then put $x \to 0$ you need not worry about other terms in Nr as they have positive powers of x and become 0 
so finally u are left with $e^{{((\ln a)^2)-((\ln b)^2)}/2}$ which is the reqd limit
A: 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)$.
A: 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 $$L = \exp\left(\frac{(\log a)^{2} - (\log b)^{2}}{2}\right)$$
