Question

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

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

Answer 1

The correct option is B ${(n!)}^{\frac{1}{n}}$

$\underset{x\to 0}{lim}{\left(\frac{1^x+2^x+....+n^x}{n}\right)}^{\frac{1}{x}}\left[1^{\infty }\text{form}\right]$
$=e^{\underset{x\to 0}{lim}(\frac{1^x+2^x+....+n^x}{n}-1)\frac{1}{x}}$
$=e^{\underset{x\to 0}{lim}\left(\frac{1^x+2^x+....+n^x-n}{x}\right)\frac{1}{n}}$
$=e^{\underset{x\to 0}{lim}(\frac{1^x-1}{x}+\frac{2^x-1}{x}+\frac{3^x-1}{x}+\cdots +\frac{n^x-1}{x})\frac{1}{n}}$

$(\because \underset{x\to 0}{lim}\frac{a^x-1}{x}=\ln a)$
$=e^{(\ln 1+\ln 2+\ln 3+\cdots +\ln n)\frac{1}{n}}$

$=e^{\frac{1}{n}\ln (n!)}$
$=e^{\ln \left(\right(n!{)}^{1/n})}$
$=(n!{)}^{1/n}$