How to show that for $\vert x \vert \lt 1$, $\lim \limits_{n \to \infty} x^n = 0$? I have tried to use the monotone convergence theorem, but that still runs into the same problem as trying to prove it with a normal $N$ and $\epsilon$ argument, which I somehow cannot figure out how to do, especially when we have $x = 0.999$ or something like that.
 A: Notice the sequence $|x|^{n}$ is decreasing when $|x| < 1$. Indeed, one has that
\begin{align*}
|x| < 1 \Rightarrow |x|^{2} < |x| < 1 \Rightarrow |x|^{3} < |x|^{2} < |x| < 1 \Rightarrow \ldots
\end{align*}
Moreover, $|x|^{n}$ is bounded below by zero. Hence it converges.
Once $|x|\xrightarrow{n\to\infty}|x|$ (because it is a constant) and every subsequence converges to the same limit, one has
\begin{align*}
L = \lim_{n\to\infty}|x|^{n+1} = \lim_{n\to\infty}|x|\times|x|^{n} = \lim_{n\to\infty}|x|\times\lim_{n\to\infty}|x|^{n} = |x|\times L \Rightarrow L(1 - |x|) = 0 \Rightarrow L = 0
\end{align*}
At last but not least, one has that
\begin{align*}
\lim_{n\to\infty}|x|^{n} = \lim_{n\to\infty}|x^{n}| = 0 \Rightarrow \lim_{n\to\infty} x^{n} = 0
\end{align*}
and we are done
A: The definitios goes like :
$$\forall \epsilon >0 \exists N : n \geq N \implies |x^n| = |x|^n < \epsilon $$
So for any arbitrary $\epsilon$ you have
$$|x|^n < \epsilon \iff n > \frac{\ln \epsilon}{\ln |x|}$$
then you can take $N = \newcommand{\ceil}[1]{\left\lceil #1 \right\rceil} \ceil{\frac{\ln \epsilon}{\ln |x|}}$.
A: Suppose $0 < x < 1$ and set $x_1 := \frac{1}{x} > 1$. Let $\varepsilon > 0$ be arbitrary and set $K := \frac{1}{\varepsilon}$. Then you can find a $n \in \mathbb{N}$ such that
$$
x_1^n > \frac{1}{\varepsilon} \iff x^n < \varepsilon,
$$
meaning that $\lim\limits_{n \to \infty} x^n = 0$. The case $-1 < x <0$ is proved analogously,
A: To do this at the very beginning of the course, you can use Bernoulli's inequality.  If $y>0$, then $(1+y)^n > 1+ny$.  Prove it by induction.  Then use this to show:
If $z>1$, then $z^n \to \infty$.  Then use this to show:
If $0<x<1$, then $x^n \to 0$.  Finally, $|-x| = |x|$ to get the case $-1<x<0$.
