\(\lim _{x\to 0+}\left(\frac{\cot^{-1} \left(\ln \left(\left\{x\right\

1.	\(\lim _{x\to 0+}\left(\frac{\cot^{-1} \left(\ln \left(\left\{x\right\}\right)\right)}{\pi }\right)^{-\ln \left(\left\{x\right\}\right)}\)is equal toA. \(e^{\frac{1}{\pi }}\)B. \(e^{-\frac{1}{\pi }}\)C. \(e^{-\frac{2}{\pi }}\)D. \(e^{\frac{2}{\pi }}\)
Answer» \(\lim\limits_{x\to 0^+}(\frac{cot^{-1}(ln x)}{π})^{-lnx}\) (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, \(\lim\limits_{x\to 0^+}(\frac{cot^{-1}(ln x)}{π})^{-lnx}\) = \(e^{1/π}\)

\(\lim _{x\to 0+}\left(\frac{\cot^{-1} \left(\ln \left(\left\{x\right\}\right)\right)}{\pi }\right)^{-\ln \left(\left\{x\right\}\right)}\)is equal toA. \(e^{\frac{1}{\pi }}\)B. \(e^{-\frac{1}{\pi }}\)C. \(e^{-\frac{2}{\pi }}\)D. \(e^{\frac{2}{\pi }}\)

Answer»

\(\lim\limits_{x\to 0^+}(\frac{cot^{-1}(ln x)}{π})^{-lnx}\) (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, \(\lim\limits_{x\to 0^+}(\frac{cot^{-1}(ln x)}{π})^{-lnx}\) = \(e^{1/π}\)