Question

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

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

Answer 1

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

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

$=\underset{x\to 0}{lim}{[1+(\frac{1^x+2^x+3^x+\dots +n^x}{n}-1)]}^{\frac{1}{x}}$

$=\underset{x\to 0}{lim}{[1+\left(\frac{1^x+2^x+3^x+\dots +n^x-n}{n}\right)]}^{\frac{1}{x}}$

$=e^{\underset{x\to 0}{lim}\left(\frac{1^x+2^x+3^x+\dots +n^x-n}{nx}\right)}$

$=e^{\frac{1}{n}\underset{x\to 0}{lim}[\frac{(1^x-1)}{x}+\frac{(2^x-1)}{x}+\frac{(3^x-1)}{x}+\dots +\frac{(n^x-1)}{x}]}$

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

$=e^{\frac{1}{n}\left(\log \right(1\cdot 2\cdot 3\cdot ..n\left)\right)}=e^{\frac{1}{n}\log (n!)}=(n!{)}^{1/n}$