<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] 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 $e^{(\ln a)^2/2 -(\ln b)^2/2}.$
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