Question

$\underset {x \rightarrow \pi} { \text{Lim}}\frac{x^\pi-\pi^x}{x^x-\pi^\pi}$ 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
$l =\underset {x \rightarrow \pi} { \text{Lim}} \frac{\pi x^{\pi-1}-\pi^x \ln \pi}{x^x(1+\ln x)}=\frac{\pi^\pi-\pi^\pi \ln \pi}{\pi^\pi(1+\ln x)}=\frac{1-\ln \pi}{1+\ln \pi} \Rightarrow \text { (A) }$
$\text { also } l =\tan \left(\tan ^{-1}(1)-\tan ^{-1}(\ln \pi)\right) \Rightarrow \text { (D) } $
$=\tan \left(\frac{\pi}{2}-\cot ^{-1} 1-\tan ^{-1}(\ln \pi)\right)=\tan \left(\cot ^{-1}(\ln \pi)-\cot ^{-1}(1)\right) \Rightarrow (D)$