Proof that $\lim\limits_{n \to \infty} (1-\frac{2}{n})^n$ exists I know the proof that $\lim\limits_{n \to \infty} (1+\frac{1}{n})^n$ exists, but I don't think I can use the same proof using the binomial theorem for this because I get $2^i$ in there and that increases without bound. Is there another method for proving this?
To show that the sequence $a_n=\left(1-\frac2n\right)^n$ converges, we will show that it is monotonically increasing and bounded above.

To show that it is monotonically increasing we analyze the ratio $\frac{a_{n+1}}{a_n}$.  To that end, we write for $n\ge 2$

$$\begin{align}
\frac{a_{n+1}}{a_n}&amp;=\frac{\left(1-\frac2{n+1}\right)^{n+1}}{\left(1-\frac2n\right)^n}\\\\
&amp;=\left(1+\frac{2}{(n-2)(n+1)}\right)^{n+1}\left(1-\frac2n\right) \tag 1\\\\
&amp;\ge \left(1+\frac{2}{n-2}\right)\left(1-\frac2n\right) \tag 2\\\\
&amp;=1
\end{align}$$

Thus, $a_{n+1}\ge a_n$ for $n\ge 2$.  In going from $(1)$ to $(2)$, we made use of Bernoulli's Inequality.

Now, it is obvious that $a_n=\left(1-\frac2n\right)^n\le 1$.  And finaly, since $a_n$ is monotonically increasing and bounded above by $1$, the sequence converges.
$$\lim_{n\to \infty}\left(1-\frac{2}{n}\right)^n=\lim_{n\to \infty}\frac{1}{\left(\left(1+\frac{1}{\frac{n-2}{2}}\right)^{\frac{n-2}{2}}\right)^{\frac{2n}{n-2}}}$$

$$=\frac{1}{e^{\lim_{n\to \infty}\frac{2n}{n-2}}}=\frac{1}{e^{\lim_{n\to \infty}\left(2+\frac{4}{n-2}\right)}}=e^{-2}$$