Question
$$\mathop {\lim }\limits_{n \to \infty } {\left\{ {\frac{{n!}}{{{{\left( {kn} \right)}^n}}}} \right\}^{\frac{1}{n}}},$$ where $$k \ne 0$$ is a constant and $$n\, \in \,N,$$ is equal to :
A.
$$ke$$
B.
$${k^{ - 1}}.e$$
C.
$$k{e^{ - 1}}$$
D.
$${k^{ - 1}}.{e^{ - 1}}$$
Answer :
$${k^{ - 1}}.{e^{ - 1}}$$
Solution :
$$\eqalign{
& {\text{Limit}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{k}{\left\{ {\frac{1}{n}.\frac{2}{n}.\frac{3}{n}......\frac{n}{n}} \right\}^{\frac{1}{n}}} \cr
& = \frac{1}{k}{e^{\mathop {\lim }\limits_{n \to \infty } \,\frac{1}{n}\,\sum\limits_{r = 1}^n {\,\log \frac{r}{n}} }} \cr
& = \frac{1}{k}{e^{\int_0^1 {\log \,x\,dx} }} \cr
& = \frac{1}{k}{e^{\left[ {x\,\log \,x} \right]_0^1 - \int_0^1 {x.\frac{1}{x}dx} }} \cr
& = \frac{1}{k}{e^{ - 1}} \cr
& \left( {\because \mathop {\lim }\limits_{x \to 0} \,x\log \,x = \mathop {\lim }\limits_{x \to 0} \frac{{\log \,x}}{{\frac{1}{x}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{x}}}{{ - \frac{1}{{{x^2}}}}} = \mathop {\lim }\limits_{x \to 0} \left( { - x} \right) = 0} \right) \cr} $$