Question

The value of $\underset{x\to \frac{\pi }{2}}{lim}{\left(\frac{1^{{\cos }^{2}x}+2^{{\cos }^{2}x}+...+n^{{\cos }^{2}x}}{n}\right)}^{\frac{1}{{\cos }^{2}x}}$ is

• A. $\frac{1}{{(n!)}^n}$
• B. ${(n!)}^{1/n}$
• C. $\frac{n^n}{n!}$
• D. ${\left(\frac{n!}{1-n}\right)}^{\frac{n}{n+1}}$

Answer 1

Answer: (B)

${(n!)}^{1/n}$

Let $f\left(x\right)=\frac{1^{{\cos }^{2}x}+2^{{\cos }^{2}x}+...+n^{{\cos }^{2}x}}{n}.$

Then, $\underset{x\to \frac{\pi }{2}}{lim}f\left(x\right)=\frac{1^0+2^0+...+n^0}{n}=\frac{1+1+...n}{n}=\frac{n}{n}=1.$

$\therefore \underset{x\to \frac{\pi }{2}}{lim}{\left[f\left(x\right)\right]}^{g\left(x\right)}=e^{{lim}_{x\to \frac{\pi }{2}}[f\left(x\right)-1]g\left(x\right)}$

using above identity we get,

$\therefore \underset{x\to \frac{\pi }{2}}{lim}{\left[\frac{1^{{\cos }^{2}x}+2^{{\cos }^{2}x}+...+n^{{\cos }^{2}x}}{n}\right]}^{\frac{1}{{\cos }^{2}x}}=e^{{lim}_{x\to \frac{\pi }{2}}\left[\frac{1^{{\cos }^{2}x}+2^{{\cos }^{2}x}+...+n^{{\cos }^{2}x}}{n}\right]\frac{1}{{\cos }^{2}x}}$

$=e^{{lim}_{x\to \frac{\pi }{2}}\left[\frac{(1^{{\cos }^{2}x}+2^{{\cos }^{2}x}+...+n^{{\cos }^{2}x})-n}{{\cos }^{2}x}\right]\frac{1}{n}}=e^{{lim}_{x\to \frac{\pi }{2}}\frac{1}{h}[\frac{1^{{\cos }^{2}x}-1}{{\cos }^{2}x}+\frac{2^{{\cos }^{2}x}-1}{{\cos }^{2}x}+...+\frac{n^{{\cos }^{2}x}-1}{{\cos }^{2}x}]}$

$=e^{\frac{1}{n}[{lim}_{{\cos }^{2}x\to 0}\left(\frac{1^{{\cos }^{2}x}-1}{{\cos }^{2}x}\right)+{lim}_{{\cos }^{2}x\to 0}\left(\frac{2^{{\cos }^{2}x}-1}{{\cos }^{2}x}\right)+{lim}_{{\cos }^{2}x\to 0}\left(\frac{n^{{\cos }^{2}x}-1}{{\cos }^{2}x}\right)]}$

$=e^{\frac{1}{n}(\log 1+\log 2+...+\log n)}=e^{\frac{1}{n}\log \left(\mathrm{1.2.3...}n\right)}=e^{\log {\left(n\right)}^{1/n}}={(n!)}^{1/n}$