Question

The value of $\underset{x\to \infty }{\lim }{\left[\frac{1^{1/x}+2^{1/x}+3^{1/x}+\dots +n^{1/x}}{n}\right]}^{nx}$ is

• A. $n!$
• B. $n$
• C. $(n-1)!$
• D. $0$

Answer 1

Answer: (A)

$\underset{x\to \infty }{\lim }{\left(\frac{1^{1/x}+2^{1/x}+3^{1/x}+\dots +n^{1/x}}{n}\right)}^{nx}$
$=\underset{y\to 0}{\lim }{\left(\frac{1^y+2^y+3^y+\dots +n^y}{n}\right)}^{\frac{n}{y}}$
$=e^{\underset{y\to 0}{\lim }\frac{n}{y}}\left(\frac{1^y+2^y+3^y+\dots +n^y}{n}-1\right)$
$=e^{\underset{y\to 0}{\lim }\left(\frac{1^y+2^y+3^y+\dots +n^y-n}{y}\right)}$
$=e^{\underset{y\to 0}{\lim }\left[\frac{\left(1^y-1\right)}{y}+\frac{\left(2^y-1\right)}{y}+\frac{\left(3^y-1\right)}{y}+\dots +\frac{\left(n^y-1\right)}{y}\right]}$
$=e^{(\log 1+\log 2+\log 3+\dots +\log n)}$
$=e^{\log (1\cdot 2\cdot 3\dots n)}=n!$