Question

$\underset{x\to \pi }{\mathrm{Lim}}\frac{x^{\pi }-{\pi }^x}{x^x-{\pi }^{\pi }}$ is equal to

• A. ${\log }_{\pi e}\left(\frac{e}{\pi }\right)$
• B. ${\log }_{\pi e}\left(\frac{\pi }{e}\right)$
• C. $\tan \left({\cot }^{-1}(\ln \pi )-{\cot }^{-1}(1)\right)$
• D. $\tan \left({\tan }^{-1}(1)-{\tan }^{-1}(\ln \pi )\right)$

Answer 1

Answer: (A), (C), (D)

Using L'Hospital's Rule
$l=\underset{x\to \pi }{\mathrm{Lim}}\frac{\pi x^{\pi -1}-{\pi }^x\ln \pi }{x^x(1+\ln x)}=\frac{{\pi }^{\pi }-{\pi }^{\pi }\ln \pi }{{\pi }^{\pi }(1+\ln x)}=\frac{1-\ln \pi }{1+\ln \pi }\Rightarrow \text{ (A) }$
$\text{ also }l=\tan \left({\tan }^{-1}(1)-{\tan }^{-1}(\ln \pi )\right)\Rightarrow \text{ (D) }$
$=\tan \left(\frac{\pi }{2}-{\cot }^{-1}1-{\tan }^{-1}(\ln \pi )\right)=\tan \left({\cot }^{-1}(\ln \pi )-{\cot }^{-1}(1)\right)\Rightarrow (D)$