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.

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}
$
