Evaluate $\lim_{n\to \infty}(0.9999+\frac{1}{n})^n$ $\lim_{n\to \infty}(0.9999+\frac{1}{n})^n$
Using Binomial theorem:
$(0.9999+\frac{1}{n})^n={n \choose 0}*0.9999^n+{n \choose 1}*0.9999^{n-1}*\frac{1}{n}+{n \choose 2}*0.9999^{n-2}*(\frac{1}{n})^2+...+{n \choose n-1}*0.9999*(\frac{1}{n})^{n-1}+{n \choose n}*(\frac{1}{n})^n=0.9999^n+0.9999^{n-1}+\frac{n-1}{2n}*0.9999^{n-2}+...+n*0.9999*(\frac{1}{n})^{n-1}+(\frac{1}{n})^n$
A limit of each element presented above is 0.  How should I prove that limit of "invisible" elements (I mean elements in "+..+") is also 0?

Hint Look at $n\gt 10000$

 For $n\gt 10000$, we have $0.9999 + \frac{1}{n} \leq 0.9999+ \frac{1}{10001}\lt1$, so $\lim_{n\rightarrow\infty}(0.9999 + \frac{1}{n})^n \leq \lim_{n\rightarrow\infty} (0.9999 + \frac{1}{10001})^n = 0$

