Question

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

• A. ${\log }_{\pi e}\left(\frac{e}{\pi }\right)$
• B. ${\log }_{\pi e}\left(\frac{\pi }{e}\right)$
• C. ${\log }_e\frac{e}{\pi }\times {\log }_{e\pi }e$
• D. ${\log }_e\frac{1}{\pi }\times {\log }_{e\pi }e$

Answer 1

The correct option is C ${\log }_e\frac{e}{\pi }\times {\log }_{e\pi }e$

We have,

${lim}_{x\to \pi }\left(\frac{x^{\pi }-{\pi }^x}{x^x-{\pi }^{\pi }}\right)$

This is the $\frac{0}{0}$ form.

So, apply L-Hospital rule,

${lim}_{x\to \pi }\left(\frac{\pi x^{\pi -1}-{\pi }^x{\log }_e\pi }{x^x({\log }_{e}x+1)-0}\right)$

${lim}_{x\to \pi }\left(\frac{\pi x^{\pi -1}-{\pi }^x{\log }_e\pi }{x^x({\log }_{e}x+1)}\right)$

$=\frac{\pi {\pi }^{\pi -1}-{\pi }^{\pi }{\log }_e\pi }{{\pi }^{\pi }({\log }_e\pi +1)}$

$=\frac{{\pi }^{\pi }-{\pi }^{\pi }{\log }_e\pi }{{\pi }^{\pi }({\log }_e\pi +1)}$

$=\frac{1-{\log }_e\pi }{({\log }_e\pi +1)}$

$=\frac{{\log }_{e}e-{\log }_e\pi }{({\log }_e\pi +{\log }_{e}e)}$

$=\frac{{\log }_e\frac{e}{\pi }}{{\log }_e\left(e\pi \right)}$

$={\log }_e\frac{e}{\pi }\times {\log }_{e\pi }e$

Hence, this is the answer.