binomial limit when limit approches to infinity $\displaystyle \lim_{n\rightarrow\infty}\binom{n}{x}\left(\frac{m}{n}\right)^x\left(1-\frac{m}{n}\right)^{n-x}$
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
 A: 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}
A: 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}
A: 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 
$$\lim_{n\to \infty}\binom{n}{x}\left(\frac{m}{n}\right)^x\left(1-\frac mn\right)^{n-x}=\frac{m^x}{x!}e^{-m}$$
A: 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 $$\infty \times e^{-m} \times 1 =\infty$$
