$$\mathop {{\text{Lim}}}\limits_{n \to \infty } {\left\{ {\frac{{n!}}{{{{\left( {kn} \right)}^n}}}} \right\}^{\frac{1}{n}}},$$     where $$k \ne 0$$  is a constant and $$n\, \in \,{\bf{N}}$$  is equal to :

Solution :
Let $$P = \mathop {{\text{Lim}}}\limits_{n \to \infty } {\left\{ {\frac{{n!}}{{{{\left( {kn} \right)}^n}}}} \right\}^{\frac{1}{n}}}$$
Taking $$\log $$  of both the sides at the base $$e$$
$$\eqalign{ & {\log _e}P = \mathop {{\text{Lim}}}\limits_{n \to \infty } \frac{1}{n}{\log _e}\left\{ {\frac{{n!}}{{{{\left( {kn} \right)}^n}}}} \right\} \cr & = \mathop {{\text{Lim}}}\limits_{n \to \infty } \frac{1}{n}{\log _e}\left\{ {\frac{1}{{kn}}.\frac{2}{{kn}}.\frac{3}{{kn}}......\frac{n}{{kn}}} \right\} \cr & = \mathop {{\text{Lim}}}\limits_{n \to \infty } \frac{1}{n}\left[ {\log \left( {\frac{1}{{kn}}} \right) + \log \left( {\frac{2}{{kn}}} \right) + ...... + \log \left( {\frac{n}{{kn}}} \right)} \right] \cr & = \mathop {{\text{Lim}}}\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^n {\log \left( {\frac{r}{{kn}}} \right)} \cr & = \int\limits_0^1 {\log } \left( {\frac{x}{k}} \right)dx \cr & = \int\limits_0^1 {\left( {\log \,x - \log \,k} \right)dx} \cr & = \int\limits_0^1 {\log } \,x\,dx - \int\limits_0^1 {\log \,k\,dx} \cr & = \left[ {x\,\log \,x - x} \right]_0^1 - \log \,k\left[ x \right]_0^1 \cr & = \left[ {0 - 1 - 0 + 0} \right] - \log \,k \cr & = - 1 - \log \,k \cr & = - \left( {\log \,e + \log \,k} \right) \cr & = - \log \left( {ek} \right) \cr & = \log \frac{1}{{ek}} \cr & \left[ {{\text{Value of}}\,x\,\log \,x\,{\text{at}}\,x = 0\,{\text{is}}\,\mathop {{\text{Lim}}}\limits_{x \to {0^ + }} x\,\log \,x = 0} \right] \cr & \therefore \,P = \frac{1}{{ek}} = {k^{ - 1}}{e^{ - 1}} \cr} $$