binomial limit when limit approches to infinity

Question

$\displaystyle \lim_{n\rightarrow\infty}\binom{n}{x}\left(\frac{m}{n}\right)^x\left(1-\frac{m}{n}\right)^{n-x}$

solution i try $\displaystyle \lim_{n\rightarrow\infty}\left(\frac{m}{n}\right)^x\left(1-\frac{m}{n}\right)^{n-x}=\lim_{n\rightarrow\infty}(\frac{m}{n}+1-\frac{m}{n})^n=1$

I have edited my post

This is wrong how i find right answer. Help me

Answer 1

If $x=0$, the limit is $\exp(-m)$.

if $x>0$, \begin{align} \lim_{n \to \infty} \left( \frac{m}{n}\right)^x \left( 1-\frac{m}n\right)^{n-x} &=\lim_{n \to \infty}\left(\frac{m/n}{1-m/n}\right)^x \lim_{n \to \infty} \left( 1 - \frac{m}{n}\right)^n\\ &=0\cdot e^{-m} \\ &=0 \end{align}

Edit to answer the edited question:

\begin{align} \lim_{n \to \infty} \binom{n}{x} \left( \frac{m}{n} \right)^x \left( 1-\frac{m}{n}\right)^{n-x} &= \lim_{n \to \infty} \frac{n!}{x!(n-x)!} \frac{m^x}{n^x} \left( 1-\frac{m}{n}\right)^{n-x} \\ &=\lim_{n \to \infty} \frac{n!}{(n-x)!n^x} \frac{m^x}{x!}\left( 1- \frac{m}{n} \right)^{n-x} \\ &=\frac{m^x}{x!} \lim_{n \to \infty} \left(1-\frac{m}n \right)^{n}\\ &=\frac{m^x}{x!} \exp(-m) \end{align}

Answer 2

Assuming that $x>0$

$$\displaystyle \lim_{n\rightarrow\infty}\left(\frac{m}{n}\right)^x\left(1-\frac{m}{n}\right)^{n-x}$$

$$\displaystyle \lim_{n\rightarrow\infty}\left(\frac{m}{n}\right)^x\left(1-\frac{m}{n}\right)^{n} \left(1-\frac{m}{n}\right)^{-x}=$$

$$0\times e^{-m} \times 1 =0$$

Assuming $x=0$ the limit is $$1\times e^{-m} \times 1 =e^{-m}$$

Assuming $x<0$ the limit is $$\infty \times e^{-m} \times 1 =\infty$$

Answer 3

Assuming $x$ is a natural number less than $n$, then we see that \begin{align} \binom{n}{x} \left(\frac{m}{n}\right)^x\left(1-\frac{m}{n}\right)^{n-x} =&\ \frac{n!}{x!(n-x)!}\left(\frac{m}{n}\right)^x\left(1-\frac{m}{n}\right)^{n-x}\\ \text{Stirling Approximation } \sim&\ \frac{\sqrt{2\pi n} \left(\frac{n}{e} \right)^n}{x! \sqrt{2\pi(n-x)}(\frac{n-x}{e})^{n-x}}\left(\frac{m}{n}\right)^x\left(1-\frac{m}{n}\right)^{n-x}\\ =&\ \frac{1}{x!}\sqrt{\frac{n}{n-x}}\left(1-\frac{m-x}{n-x} \right)^{n-x}\left( \frac{m}{e}\right)^x\\ \rightarrow &\ \frac{1}{x!} \left( \frac{m}{e}\right)^xe^{-(m-x)}= \frac{m^x}{x!}e^{-m} \end{align}

Answer 4

Note that we have

$$\binom{n}{x}\left(\frac{m}{n}\right)^x\left(1-\frac mn\right)^{n-x}=\frac{m^x}{x!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{x-1}{n}\right)\left(1-\frac mn\right)^{n-x}$$

Therefore, for fixed $x$, we have

$$\lim_{n\to \infty}\binom{n}{x}\left(\frac{m}{n}\right)^x\left(1-\frac mn\right)^{n-x}=\frac{m^x}{x!}e^{-m}$$