Correct Answer - Option 3 :
\(\rm \frac{1}{ke}\)
Concept:
Definite integral as the limit of a sum:
\(\rm \int_{a}^{b}{f(x)dx}=\lim_{n\to \infty }\sum\limits_{r=1}^{n}{f\left( {{x}_{r}} \right)}\Delta x \)
⇒ \(\rm \int_{a}^{b}{f(x)dx}=\lim_{n\to \infty } \sum\limits_{r=1}^{n}{f\left( a+\dfrac{(b-a)}{n}r \right)}\dfrac{(b-a)}{n}\)
Calculation:
Let's say that y = \(\rm \lim_{n\to\infty} \frac{(n!)^{\frac1n}}{kn},k\ne 0\).
Using the definition of factorials:
⇒ y = \(\rm \lim_{n\to \infty } {{\left\{ \frac{n}{kn}\times \frac{n-1}{kn}\times \frac{n-2}{kn}\times ...\times \frac{1}{kn} \right\}}^{\frac{1}{n}}}\)
Taking log of both sides, we get:
⇒ log y = \(\rm \lim_{n\to \infty } \frac{1}{n}\left\{ \log \left( \frac{n}{kn} \right)+\log \left( \frac{n-1}{kn} \right)+\log \left( \frac{n-2}{kn} \right)+...+\log \left( \frac{1}{kn} \right) \right\}\)
Which can also be written as:
⇒ log y = \(\rm \lim_{n\to \infty } \frac{1}{n}\left\{ \log \left( \frac{1}{k} \right)+\log \left( \frac{1}{k}-\frac{1}{kn} \right)+\log \left( \frac{1}{k}-\frac{2}{kn} \right)+...+\log \left( \frac{1}{k}-\frac{n-1}{kn} \right) \right\}\)
In summation form, this is equal to:
⇒ log y = \(\rm \lim_{n\to \infty } \sum\limits_{r=0}^{n-1}{\log \left( \frac{1}{k}-\frac{r}{kn} \right)}\frac{1}{n}\)
Let's say \(\rm \frac{1}{k}=a\).
⇒ log y = \(\rm \lim_{n\to \infty } \sum\limits_{r=0}^{n-1}{\log \left( a-\frac{ar}{n} \right)}\frac{1}{n}\)
⇒ log y = \(\rm \frac{1}{(0-a)}\lim_{n\to \infty } \sum\limits_{r=0}^{n-1}{\log \left( a+\frac{(0-a)r}{n} \right)}\frac{(0-a)}{n}\)
Using the integral as limit of sum \(\rm \int_{a}^{b}{f(x)dx}=\lim_{n\to \infty } \sum\limits_{r=1}^{n}{f\left( a+\frac{(b-a)}{n}r \right)}\frac{(b-a)}{n}\), we can write it as:
⇒ log y = \(\rm \frac{-1}{a}\int_{a}^{0}{\log (x)dx}\)
Using \(\rm \int_{a}^{b}{f(x)dx}=-\int_{b}^{a}{f(x)dx}\), we can re-write it as:
⇒ log y = \(\rm \frac{1}{a}\int_{0}^{a}{\log (x)dx}\)
Using \(\rm \int{\log xdx}=x\log x-x+C\), we get:
⇒ log y = \(\rm \frac{1}{a}\left[ x\log x-x \right]_{0}^{a}\)
⇒ log y = \(\rm \frac{1}{a}(a\log a-a)\)
⇒ log y = log a - 1
⇒ log y = log a - log e
Substituting back \(\rm \frac{1}{k}=a\), we get:
⇒ y = \(\rm \frac{1}{ke}\)
