Question

The value of $\displaystyle\lim _{x \rightarrow \infty}\left[\frac{1^{1 / x}+2^{1 / x}+3^{1 / x}+\ldots+n^{1 / x}}{n}\right]^{n x}$ is

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

Answer 1

Answer: (A)

$\displaystyle\lim _{x \rightarrow \infty}\left(\frac{1^{1 / x}+2^{1 / x}+3^{1 / x}+\ldots+n^{1 / x}}{n}\right)^{n x}$
$=\displaystyle\lim _{y \rightarrow 0}\left(\frac{1^{y}+2^{y}+3^{y}+\ldots+n^{y}}{n}\right)^{\frac{n}{y}}$
$=e^{\displaystyle\lim _{y \rightarrow 0} \frac{n}{y}}\left(\frac{1^{y}+2^{y}+3^{y}+\ldots+n^{y}}{n}-1\right)$
$=e^{\displaystyle\lim _{y \rightarrow 0}\left(\frac{1^{y}+2^{y}+3^{y}+\ldots+n^{y}-n}{y}\right)}$
$=e^{\displaystyle\lim _{y \rightarrow 0}\left[\frac{\left(1^{y}-1\right)}{y}+\frac{\left(2^{y}-1\right)}{y}+\frac{\left(3^{y}-1\right)}{y}+\ldots+\frac{\left(n^{y}-1\right)}{y}\right]}$
$=e^{(\log 1+\log 2+\log 3+\ldots+\log n)}$
$=e^{\log (1 \cdot 2 \cdot 3 \ldots n)}=n !$