<s>[INST] 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
 [/INST] 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}
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