<s>[INST] How find this limit $I=\lim_{x\to 0}\left(\frac{a^x-x\ln{a}}{b^x-x\ln{b}}\right)^{\cot^2{x}}$ Find the limit of the value 

$$I=\lim_{x\to 0}\left(\dfrac{a^x-x\ln{a}}{b^x-x\ln{b}}\right)^{\cot^2{x}},(0<a\neq1,0<b\neq 1,a\neq b)$$

Following is My ugly methods

$$I=e^{\displaystyle\lim_{x\to 0}\cot^2{x}\ln{\left(\dfrac{a^x-x\ln{a}}{b^x-x\ln{b}}\right)}}$$
  so we only find following this limit
  $$A=\lim_{x\to 0}\cot^2{x}\ln{\left(\dfrac{a^x-x\ln{a}}{b^x-x\ln{b}}\right)}=\lim_{x\to 0}\dfrac{\ln{(a^x-x\ln{a})}-\ln{(b^x-x\ln{b})}}{x^2}$$

By the  L'Hospital
$$A=\lim_{x\to 0}\dfrac{\dfrac{a^x\ln{a}-\ln{a}}{a^x-x\ln{a}}-\dfrac{b^x\ln{b}-\ln{b}}{b^x-x\ln{b}}}{2x}$$
so

\begin{align*}A&=\lim_{x\to 0}\dfrac{a^xb^x(\ln{a}-\ln{b})+\ln{a}\ln{b}x(b^x-a^x)+a^x\ln{b}-b^x\ln{a}}{(a^x-x\ln{a})(b^x-x\ln{b})(2x)}\\
&=\lim_{x\to 0}\dfrac{(\ln{a}-\ln{b})+\ln{a}\ln{b}x(b^x-a^x)+(\ln{b}-\ln{a})}{2x}\\
&=\lim_{x\to 0}\dfrac{\ln{a}\ln{b} x(b^x-a^x)}{2x}\\
&=0
\end{align*}

so $$I=e^{0}=1$$
Have other nice methods and my result is true? Thank you
 [/INST] Since $\ln (a^x-x\ln{a}) = \ln (1+\frac{x^2}{2}(\ln{a})^2 + o(x^2)) = \frac{x^2}{2}(\ln{a})^2 + o(x^2)$, $\ln (b^x-x\ln{b}) = \frac{x^2}{2}(\ln{b})^2 + o(x^2)$, we have
$$\cot^2{x} \left(\ln (a^x-x\ln{a}) - \ln (b^x-x\ln{b}\right) = \frac{(\ln{a})^2-(\ln{b})^2}{2} + o(1).$$
Hence

$$I = \sqrt{e^{(\ln{a})^2-(\ln{b})^2}}.$$

</s>