Finding the limit of a function $n^3 / 3^n$ $$\lim_{n→\infty}  \frac{n^3}{3^n} =0 $$
The answer is 0 but how would i go about proving that?

Let $a_n = \frac{n^3}{3^n} $. Then
$$ \left| \frac{ a_{n+1}}{a_n} \right| = \frac{(n+1)^3}{3^{n+1}} \frac{3^n}{n^3} = \frac{1}{3} \left( \frac{n+1}{n} \right)^3 \to \frac{1}{3} < 1$$
Therefore $a_n \to 0 $
