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Question: \({\underset{n \rightarrow \infty}{Lim}}^{n}C_{x}\left( \frac{m}{n} \right)^{x}\left( 1 - \frac{m}{n...

Limn→∞nCx(mn)x(1−mn)n−x{\underset{n \rightarrow \infty}{Lim}}^{n}C_{x}\left( \frac{m}{n} \right)^{x}\left( 1 - \frac{m}{n} \right)^{n - x} equals to –

A

mxx!.e−m\frac{m^{x}}{x!}.e^{- m}

B

mxx!.em\frac{m^{x}}{x!}.e^{m}

C

e0

D

0

Answer

mxx!.e−m\frac{m^{x}}{x!}.e^{- m}

Explanation

Solution

Ltn→∞nxn−xmxnx×(1−mn)n(1−mn)x\underset{n \rightarrow \infty}{Lt}\frac{\begin{matrix} n \end{matrix}}{\begin{matrix} x \end{matrix}\begin{matrix} n - x \end{matrix}}\frac{m^{x}}{n^{x}} \times \frac{\left( 1 - \frac{m}{n} \right)^{n}}{\left( 1 - \frac{m}{n} \right)^{x}}

= mxxLtn→∞(1−mn)n\frac{m^{x}}{\begin{matrix} x \end{matrix}}\underset{n \rightarrow \infty}{Lt}\left( 1 - \frac{m}{n} \right)^{n} Ltn→∞nn−x(1−mn)−xnx\underset{n \rightarrow \infty}{Lt}\frac{\begin{matrix} n \end{matrix}}{\begin{matrix} n - x \end{matrix}}\frac{\left( 1 - \frac{m}{n} \right)^{- x}}{n^{x}}

= Ltn→∞nnxn−x\underset{n \rightarrow \infty}{Lt}\frac{\begin{matrix} n \end{matrix}}{n^{x}\begin{matrix} n - x \end{matrix}} Ltn→∞(1−mn)−x\underset{n \rightarrow \infty}{Lt}\left( 1 - \frac{m}{n} \right)^{- x}

= e−mmxxLtn→∞nn−xnx\frac{e^{- m}m^{x}}{\begin{matrix} x \end{matrix}}\underset{n \rightarrow \infty}{Lt}\frac{\begin{matrix} n \end{matrix}}{\begin{matrix} n - x \end{matrix}n^{x}}