Question

The value of the $\underset{x\to 0}{lim}{\left\{\frac{e^x+(2e{)}^x+(3e{)}^x+...+(ne{)}^x}{n}\right\}}^{n/x}$ is

• A. $n+lo{g}_{e}n!$
• B. $n!e^n$
• C. $n!$
• D. $e^n$

Answer 1

The correct option is B $n!e^n$

$\underset{x\to 0}{lim}{\left(\frac{e^x+2^x{e}^x+3^x.e^x+....+n^x{e}^x}{n}\right)}^{n/x}$
$y=\underset{x\to 0}{lim}(e^x{)}^{n/x}{\left(\frac{1^x+2^x+3^x+....+n^x}{n}\right)}^{n/x}$
$y=e^n\underset{x\to 0}{lim}{\left({\left(\frac{1^x+2^x+3^x+....+n^x}{n}\right)}^{1/x}\right)}^n$
We know that
$\underset{x\to 0}{lim}{\left(\frac{a_1^x+a_2^x+a_3^x+.....+a_n^x}{n}\right)}^{\frac{1}{x}}={(a_1{a}_2{a}_3.....a_n)}^{1/n}$
Hence, the given limit becomes
$=e^n{\left(\right(\mathrm{1.2.3......}n{)}^{1/n})}^n$
$=e^n\left(\right(n!{)}^{1/n}{)}^n$
$=e^n.n!$