Integral Calculus Question 144
Question: $ \underset{n\to \infty }{\mathop{Lim}},{{{ \frac{n!}{{{(kn)}^{n}}} }}^{\frac{1}{n}}}, $ where $ k\ne 0 $ is a constant and $ n\in N $ is equal to
Options:
A) ke
B) $ {k^{-1}}e $
C) $ k{e^{-1}} $
D) $ {k^{-1}}{e^{-1}} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Let $ P=\underset{n\to \infty }{\mathop{Lim}},{{{ \frac{n!}{{{(kn)}^{n}}} }}^{\frac{1}{n}}} $ Taking log of both the sides at the base e $ {\log_{e}}P=\underset{n\to \infty }{\mathop{Lim}},\frac{1}{n}{\log_{e}}{ \frac{n!}{{{(kn)}^{n}}} } $ $ =\underset{n\to \infty }{\mathop{Lim}},\frac{1}{n}{\log_{e}}{ \frac{1}{kn}.\frac{2}{kn}.\frac{3}{kn}………………\frac{n}{kn} } $ $ =\underset{n\to \infty }{\mathop{Lim}},\frac{1}{n}[ \log ( \frac{1}{kn} )+\log ( \frac{2}{kn} )+……+\log ( \frac{n}{kn} ) ] $ $ =\underset{n\to \infty }{\mathop{Lim}},\frac{1}{n}\sum\limits_{r=1}^{n}{\log ( \frac{r}{kn} )} $ $ =\int\limits_0^{1}{\log ( \frac{x}{k} )dx=\int\limits_0^{1}{(\log x-\log k)dx}} $ $ =\int\limits_0^{1}{\log x,dx-\int\limits_0^{1}{\log kdx}} $ $ =[ x\log x-x ]_0^{1}-\log k[ x ]_0^{1} $ $ =[ 0-1-0+0 ]-\log k=-1-\log k $ $ =-(\log e+\log k)=-\log (ek)=\log \frac{1}{ek} $ [Value of $ x\log xatx=0 $ is $ \underset{x\to {0^{+}}}{\mathop{Lim}},x\log x=0 $ ]
$ \therefore ,P=\frac{1}{ek}={k^{-1}}{e^{-1}} $
BETA
