Question

${lim}_{n\to {\infty }^{}}{}^n{C}_r{\left(\frac{m}{n}\right)}^r{\left(1-\frac{m}{n}\right)}^{n-r}$, where $m<n$ is equals to:

• A. $e^{-m}{m}^r$
• B. $\frac{m^r}{r!}$
• C. $\frac{m^r{e}^{-m}}{r!}$
• D. $\frac{e^{-r}{r}^m}{m!}$

Answer 1

Answer: (C)

$\frac{m^r{e}^{-m}}{r!}$

$\underset{n\to \infty }{lim}{}^n{C}_r{\left(\frac{m}{n}\right)}^r{\left(1-\frac{m}{n}\right)}^{n-r}$

$=\underset{n\to \infty }{lim}\frac{n(n-1)(n-2)...(n-r+1)}{r!}{\left(\frac{m}{n}\right)}^r{\left({\left(1-\frac{m}{n}\right)}^{-\frac{n}{m}}\right)}^{-\frac{(n-r)m}{n}}$

$=\underset{n\to \infty }{lim}\frac{n(n-1)(n-2)...(n-r+1)}{\left(n^r\right)r!}{m}^r\times \underset{n\to \infty }{lim}{\left({\left(1-\frac{m}{n}\right)}^{-\frac{n}{m}}\right)}^{-\frac{(n-r)m}{n}}$

$=\underset{n\to \infty }{lim}\frac{(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{r-1}{n})}{r!}{m}^r.e^{(1-0)(-m)}$

$=\frac{(1-0)(1-0)...(1-0)}{r!}{m}^r{e}^{-m}=\frac{e^{-m}{m}^r}{r!}$
Hence, option 'C' is correct.