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?

Wow your approach to this question is way too difficult.
Let me first explain what's wrong with your approach before I give you another way to do this.
You wrote your term as a sum of $n$ terms, each converges to zero as $n$ goes to infinity . In fact you didn't show that the "invisible" elements converges to zero, but even if you do, it is not enough. For instance the sum $\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}$ of $n$ times $\frac{1}{n}$ converges to $1$ even though each term converges to zero.
What can you do instead:
Instead you can say that for $n$ sufficiently large $0.9999 + \frac{1}{n} < 0.99999$ (add one more 9). Then use the fact that $0.99999^n\rightarrow 0$ as $n\rightarrow 0$.
