How do I prove that if $|x|<1$, then $\lim_{n\to\infty}x^n=0$? Prove that:
If $|x|<1$, then $\lim_{n\to\infty}x^n=0$.

My thought:
Since it's a theory so I'm not able to solve it mathematically also I'm too confused about "infinity algebra". I have just memorized "infinity algebra results."
So, according to me, infinity is a very large number, therefore if any number is raised to power infinity then ultimately we'll get infinity.

Answer:It's a simple algebra if you'll increase power of $\vert x  \vert<1$ then it'll keep decreasing the value hence you'll get $0$ too simple.
Edit:Thanku dear teachers for helping.I know it was too simple but sometimes we get struck on silly doubts like this.
 A: Bernoulli's inequality is all you need.
If
$0 < x < 1$
then
$x = \dfrac1{1+a}$
where
$a = \dfrac1{x}-1
\gt 0$.
Then
$(1+a)^n \ge 1+na
\gt na
=n(\dfrac1{x}-1)$
so
$\begin{array}\\
x^n
&=\dfrac1{(1+a)^n}\\
&\lt\dfrac1{n(\dfrac1{x}-1)}\\
&=\dfrac{x}{n(1-x)}\\
&\to 0
\text{ as } n \to \infty\\
\end{array}
$
A: we know that $|x| < 1$, so $|x^n| < 1$ for every $n \in \Bbb{N}$. This means that the sequnece $a_n = x^n$ is bounded, its sufficient to prove that $a_n$ is convergent and then we can find the desires limit. 
Without losing generality we can suppose that $x > 0$, for $x = 0$ the statement is true, then its easy to show that:
$$x^{n+1} \le x \cdot x^n < x^n \quad \Longrightarrow \quad a_{n+1} < a_n, \forall n \in \Bbb{N}$$
This means that $a_n$ is a monotone sequence and then $a_n$ is convergent and has limit $\ell$, and then:
$$\lim_{n\rightarrow \infty} a_n = \ell = \lim_{n\rightarrow \infty} a_{n+1}  = x \lim_{n\rightarrow \infty} a_n, \quad \Longrightarrow \quad \ell = x \cdot \ell, \quad \Longrightarrow \quad \ell = 0 $$
For $x < 0$, we can choose two distint subsequnce of $a_n$ for both even and odd power of $x$, then both of sequnces are convergent and prove is complete.
A: For $$-1<x<1$$ you can define $$x'=\frac{1}{x}$$ if $$x\neq 0$$
A: If $$-1<x<1$$ and you raise it to any power $n > 1 $ the value of it decreases.
Eg say$ x = 0.1 , x ^2  = 0.01, x^3 = 0.001$ or $x ^n = 0.1 ^ n = \frac{1}{10^n}$ and as $ n$ keeps getting bigger , the denominator approaches $\infty $ and the numerator approaches $0 $
