Why is $\lim\limits_{N\to\infty}x^{N+1}=0$, where $|x|<1$? How is this done? 
Why is $\lim\limits_{N\to\infty}x^{N+1}=0$, where $-1<x<1$?

A relatively "low-tech" way to see the limit must be zero (assuming the limit exists) is to call the limit $L$ and note that
$$
L = \lim_{N \to \infty} x^{N+1}
  = \lim_{N \to \infty} (x \cdot x^{N})
  = x \lim_{N \to \infty} x^{N}
  = xL.
$$
Subtracting and factoring, $(1 - x)L = 0$. Since $x \neq 1$ by hypothesis, it must be that $L = 0$.

To prove the limit exists without descending into $\varepsilon$-land (i.e., using convergence criteria that appear early in an elementary analysis course), note that if $0 \leq x < 1$, then the sequence $(x^{N})_{N=0}^{\infty}$ is non-increasing and bounded below by $0$, so it has a real limit $L$.
To handle the case $-1 < x < 0$, note that $-|x|^{N} \leq x^{N} \leq |x|^{N}$ for all $N \geq 0$ and apply the squeeze theorem.
