<s>[INST] Proof using the definition of the limit. Prove that $\lim_{n\to\infty}$ $\frac{n^3}{3^n}$ $=0$ using the definition of the limit.
Here is what I have so far:
$\left\lvert{\frac{n^3}{3^n} -0}\right\rvert$ = $\left\lvert{\frac{n^3}{(1+2)^n} }\right\rvert$$<\left\lvert{\frac{n^3}{1+\binom{n}{1}2+\binom{n}{2}2^2+\binom{n}{3}2^3+\binom{n}{4}2^4+\cdots}}\right\rvert$

However, I cannot think of anyway to reduce the above equation so it is less of any $\epsilon$ greater than zero.
 [/INST] I think you did something closely related to the following. 
The fifth term in the binomial expansion of $(1+2)^n$ is $\binom{n}{4}2^4$. So for $n\gt 3$ we have 
$(1+2)^n \gt 16\cdot\frac{n(n-1)(n-2)(n-3)}{4!}$. 
Note that if $n\gt 5$ then $n-1\gt n/2$ and $n-2\gt n/2$ and $n-3\ge n/2$. It follows that
$$16\cdot\frac{n(n-1)(n-2)(n-3)}{4!}\gt 16\cdot\frac{(n)(n/2)(n/2)(n/2)}{4!}=\frac{n^4}{12}.$$
Thus  if $n\ge 5$, then
$$\frac{n^3}{3^n}\lt \frac{12}{n}.$$
Now I think you can finish.
Remark: It may have been a strategic mistake to "multiply out." The expression $16\cdot\frac{n(n-1)(n-2)(n-3)}{4!}$ is easy to think about, while a multiplied out version is less easy. 
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