\(\lim\limits_{x\to 0^+}(\frac{cot^{-1}(ln x)}{π})^{-lnx}\) (1\(\infty\) - type)
= Exp{\(\lim\limits_{x\to 0^+}(\frac{cot^{-1}(ln x)}{π}-1)(-ln x)\)}
= Exp{\(\lim\limits_{x\to 0^+}(\frac{cot^{-1}(ln x)-π}{π})(-ln x)\)}
= Exp{\(\lim\limits_{x\to 0^+}\frac{-(π-cot^{-1}(ln x))}{π})(-ln x)\)}
= Exp{\(\lim\limits_{x\to 0^+}\frac{cot^{-1}(ln x)lnx}{π}\)} ( ∵ cot-1(-x) = π - cot-1x) (0 x \(\infty\) - case)
= Exp{\(\lim\limits_{x\to 0^+}\cfrac{cot^{-1}(-ln x)}{\frac{\pi}{lnx}}\)} (0/0 - case)
= Exp{\(\left\{\lim\limits_{x\to 0^+}\cfrac{\frac{-1}{1+(lnx)^2}\times\frac{-1}x}{\frac{-\pi}{(lnx)^2}\times\frac1x}\right\}\) (By using D.L.H. Rule)
= Exp{\(\lim\limits_{x\to 0^+}\frac{-(lnx)^2}{π(1+(lnx)^2)}\)}
= Exp\(\left\{\lim\limits_{x\to 0}\cfrac{-1}{\pi(\frac1{(lnx)^2}+1)}\right\}\)
= Exp {\(\frac{-1}x\)} (∵ \(\lim\limits_{x\to0^+}\frac{1}{(lnx)^2}\) = \(\frac1{(-\infty)^2}\) = \(\frac1{\infty}\) = 0)
= e-1/x
Hence, \(\lim\limits_{x\to 0^+}(\frac{cot^{-1}(ln x)}{π})^{-lnx}\) = \(e^{1/π}\)