Finding the limit of $\left( 1-\frac{1}{n} \right)^{n}$

How would one compute the following limit?

$$\lim_{n \to \infty} \left( 1 - \frac{1}{n} \right)^{n}$$

I know

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

but right there is a minus keeping that limit from being used.

Another problem I am questioning is finding the limit of

$$\frac{n!}{2n}$$

Of course, $\frac{x^n}{n!}$ has zero as a limit but here it is the opposite.

You can obtain the limit of

$$\left(1-\frac1n\right)^n$$

easily from the one you know, $\left(1 + \frac1n\right)^n \to e$, by noting

$$1 - \frac1n = \frac{n-1}{n} = \frac{1}{\left(\frac{n}{n-1}\right)} = \frac{1}{\left(1 + \frac{1}{n-1}\right)}.$$

Then you can write

$$\left(1 - \frac1n\right)^n = \left(1-\frac1n\right)\frac{1}{\left(1+\frac{1}{n-1}\right)^{n-1}},$$

where the first factor obviously converges to $1$, and the second one converges to $\frac{1}{e}$ by what you already know.

For another one use the definition of $n!$. In fact:

$$\frac{n!}{2n}>>n$$

The result of the limit is $e^{-1}$ not $e^n$. Let $y= (1-\frac{1}{n})^n$, or $\ln y= n\ln (1-\frac{1}{n})$. Since $\displaystyle\lim_{n\rightarrow\infty}n\ln (1-\frac{1}{n})=\lim_{n\rightarrow\infty}\frac{\ln (1-\frac{1}{n})}{1/n}=-1$ by L'Hospital's rule, so $y=\frac{1}{e}$. The other limit is infinity: $\lim\frac{n!}{x^n}=\lim\frac{1}{\frac{x^n}{n!}}=\infty$. In the same manner since $\lim\frac{2n}{n!}=0$, we have $\lim\frac{n!}{2n}=\infty$.