limn→∞ ncrmnr1-mnn-r equals

limn→∞Cn,rmnr1−mnn−r=limn→∞nn−1n−2.......n−r+1r.mrnr1−mnn−r=limn→∞nn−1n−2...........n−r+1n.n.n........n.mrrlimn→∞1−mn−nmn−r−mn=limn→∞1.1−1n1−2n........1−r−1nmrr×elimn→∞−m1−rn ∵ limx→∞(1+1x)x=e=1.1.........1mrr.e−m=mre−mr

limn→∞ ncrmnr1-mnn-r equals

limn→∞Cn,rmnr1−mnn−r=limn→∞nn−1n−2.......n−r+1r.mrnr1−mnn−r=limn→∞nn−1n−2...........n−r+1n.n.n........n.mrrlimn→∞1−mn−nmn−r−mn=limn→∞1.1−1n1−2n........1−r−1nmrr×elimn→∞−m1−rn ∵ limx→∞(1+1x)x=e=1.1.........1mrr.e−m=mre−mr

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$\lim_{n \to \infty}^{}^{n} c_{r} {(\frac{m}{n})}^{r} {(1 - \frac{m}{n})}^{n - r} equals$

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$e^{- m} m^{r}$

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

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

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

answer is C.

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Detailed Solution

${\begin{array}{l} \lim_{n \to \infty} C (n, r) {(\frac{m}{n})}^{r} {(1 - \frac{m}{n})}^{n - r} \\ = \lim_{n \to \infty} \frac{n (n - 1) (n - 2) ....... (n - r + 1)}{r} . \frac{m^{r}}{n^{r}} {(1 - \frac{m}{n})}^{n - r} \\ = \lim_{n \to \infty} \frac{n (n - 1) (n - 2) ........... (n - r + 1)}{n . n . n ........ n} . (\frac{m^{r}}{r}) {[{[\lim_{n \to \infty} (1 - \frac{m}{n})]}^{\frac{- n}{m}}]}^{(n - r) (\frac{- m}{n})} \\ = \lim_{n \to \infty} [1. (1 - \frac{1}{n}) (1 - \frac{2}{n}) ........ (1 - \frac{(r - 1)}{n})] (\frac{m^{r}}{r}) \times [e^{\lim_{n \to \infty} (- m) (1 - \frac{r}{n})}] ∵ \lim_{x \to \infty} {(1 + \frac{1}{x})}^{x} = e \\ = (1.1.........1) \frac{m^{r}}{r} . e^{- m} = \frac{m^{r} e^{- m}}{r} \end{array}}^{}$