Question

$\underset{x\to 0}{lim}{\left(\frac{1^x+2^x+3^x+...+n^x}{n}\right)}^{a/x}$ is equal to

• A. ${(n!)}^a$
• B. $\frac{1}{a(n!)}$
• C. ${(n!)}^{a/n}$
• D. $\frac{n}{{(n!)}^a}$

Answer 1

Answer: (C)

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

Let $f\left(x\right)=\left(\frac{1^x+2^x+3^x+...+n^x}{n}\right).$

Then $\underset{x\to 0}{lim}f\left(x\right)=\frac{1^0+2^0+3^0+...+n^0}{n}=\frac{n}{n}=1.$

$\therefore \underset{x\to 0}{lim}{\left[f\left(x\right)\right]}^{g\left(x\right)}=e^{{lim}_{x\to 0}g\left(x\right).\{f\left(x\right)-1\}}$

Now, $\underset{x\to 0}{lim}{\left(\frac{1^x+2^x+3^x+...+n^x}{n}\right)}^{a/x}=e^{{lim}_{x\to 0}\frac{a}{x}\left\{\frac{1^x+2^x+3^x+...+n^x}{n}\right\}}$

$=e^{{lim}_{x\to 0}\left\{\frac{(1^x+2^x+3^x+...+n^x)-n}{n}\right\}\frac{a}{n}}$

$=e^{\frac{a}{n}{lim}_{x\to 0}\{\frac{(1^x-1)}{x}+\frac{(2^x-1)}{x}+\frac{(3^x-1)}{x}+...+\frac{(n^x-1)}{x}\}}$

$=e^{\frac{a}{n}(\log 1+\log 2+\log 3+...+\log n)}=e^{\log {(n!)}^{a/n}}={(n!)}^{a/n}$