Q: $\large lim_{n \rightarrow \infty} (\frac{n!}{(m n)^n})^{1/n}$ (m ∈ N) is equal to
(A) $\large \frac{1}{e m}$
(B) $\large \frac{m}{e }$
(C) e m
(D) $\large \frac{e}{ m}$
Solution :
L.H.S $\large = lim_{n \rightarrow \infty} (\frac{1.2.3…..n}{(m n)^n})^{1/n}$
$\large = lim_{n \rightarrow \infty} (\frac{1.2.3…..n}{(n)^n})^{1/n} \times \frac{1}{m}$
$\large lim_{n \rightarrow \infty} \frac{1}{m}[(\frac{1}{n}) (\frac{2}{n}) (\frac{3}{n})….(\frac{n-1}{n}) (\frac{n}{n})]^{1/n}$ = S (say)
$\large ln S = lim_{n \rightarrow \infty} [ln(\frac{1}{m}) + \frac{1}{n} \Sigma_{r=1}^{n} ln(\frac{r}{n})] $
$\large ln S = – ln m + \int_{0}^{1} ln x dx $
$\large ln S = – ln m + [x ln x – x ]_{0}^{1} $
$\large ln S = – ln m + (-1)$
$\large ln S = – ln m – ln e $
$\large ln S = – ln e m $
$\large ln S = ln\frac{1}{e m} $
$\large S = \frac{1}{e m} $
Hence (A) is the correct answer.