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?
 A: 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 $
A: Apply l´Hopital's rule three times to the function $f(x)=x^3/3^x$:
$$\lim_{x\rightarrow\infty}\left(\frac{x^3}{3^x}\right)=\lim_{x\rightarrow\infty}\left(\frac{6}{(\ln 3)^3 3^x}\right)=0$$
A: Note that $n^3 \leq 2^n$ for sufficiently large $n$. Thus, we can bound $a_n = n^3/3^n$ by
$$
0 \leq \frac{n^3}{3^n} \leq \frac{2^n}{3^n}.
$$
Furthermore, $\lim_{n \to \infty} (2/3)^n = 0$ and so we have
$$
0 \leq \lim_{n \to \infty} \frac{n^3}{3^n} \leq 0.
$$
By the squeeze theorem, the limit is $0$.
A: Let's start with the intuitive fact that 
$$\lim_{n\to \infty} \ln(\frac{n^3}{3^n}) = \lim_{n\to \infty} \ln(n^3) - \ln({3^n}) = \lim_{n\to \infty} 3\ln(n) - n\ln({3})=-\infty$$
This is clear because $\ln(n)<n$, so for large enough $n$ we have the second term dominating.
Now $\ln(\cdot)$ is continuous so 
$$0=e^{-\infty}=e^{\lim_{n\to \infty} \ln(\frac{n^3}{3^n})} = e^{\ln \lim_{n\to \infty} \frac{n^3}{3^n}} = \lim_{n\to \infty} \frac{n^3}{3^n}$$
