Question

The value of $\underset{n\to \infty }{lim}{\left(\frac{n!}{(mn{)}^n}\right)}^{1/n},$ where $m\in N$ is equal to

• A. $\frac{1}{em}$
• B. $\frac{m}{e}$
• C. $em$
• D. $\frac{e}{m}$

Answer 1

The correct option is A $\frac{1}{em}$

Let $P=\underset{n\to \infty }{lim}{\left(\frac{n!}{(mn{)}^n}\right)}^{1/n}$
$\Rightarrow P=\underset{n\to \infty }{lim}{\left(\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdots (n-1)\cdot n}{n^n}\right)}^{1/n}\times \frac{1}{m}\Rightarrow P=\underset{n\to \infty }{lim}{[\left(\frac{1}{n}\right)\cdot \left(\frac{2}{n}\right)\cdot \left(\frac{3}{n}\right)\cdot \cdots \left(\frac{n-1}{n}\right)\cdot \left(\frac{n}{n}\right)]}^{1/n}\times \frac{1}{m}$
Taking $\ln$ both sides,
$\Rightarrow \ln P=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\ln \left(\frac{r}{n}\right)+\ln \left(\frac{1}{m}\right)=\underset{0}{\overset{1}{\int }}\ln xdx-\ln m$
$=[x\ln x-x{]}_0^1-\ln m=(0-1)-(\underset{x\to 0}{lim}\frac{\ln x}{1/x}-0)-\ln m=-1-\underset{x\to 0}{lim}\frac{1/x}{-1/x^2}-\ln m=-1-\ln m$

$\Rightarrow \ln P=\ln \left(\frac{1}{em}\right)\therefore P=\frac{1}{em}$