Limit: $\lim_{n\to \infty} \frac{n^5}{3^n}$

I need help on a homework assignment. How to show that $\lim_{n\to\infty} \left(\dfrac{n^5}{3^n}\right) = 0$? We've been trying some things but we can't seem to find the answer.

Depending on what have you studied and what are you supposed to know, you might be able to use one of the following tricks (we put $a_n:=\frac{n^5}{a^n}$):

1. Show that $\lim\limits_{n\to\infty} \frac{a_n}{a_{n-1}} < 1$. (This is one of the basic tricks: If this limit is $q<1$, then eventually $\frac{a_n}{a_{n-1}}<q+\varepsilon$ for some small $\varepsilon$, and $a_n < C (q+\varepsilon)^n$, now you can use the sandwich theorem.)

2. Show that $\lim\limits_{n\to\infty} \log a_n = -\infty$. (Then $\lim\limits_{n\to\infty} a_n=0$ follows from the continuity of $\log a$ for $a>0$.)

3. L'Hospital criterion: Since both $f(x):=x^5$ and $g(x):=3^x$ are 5 times differentiable and have limit $+\infty$ up to $4$th derivative, you have $$\lim\limits_{n\to\infty} \frac{a_n}{a_{n-1}} = \lim\limits_{n\to\infty} \frac{f'(x)}{g'(x)} = \cdots = \lim\limits_{n\to\infty}\frac{f^{(5)}(x)}{g^{(5)}(x)}.$$ It remains to show that the $5$th derivative of $f$ is constant and the $5$th derivative of $g$ has limit $+\infty$.

First show that $3^n>n^6$ for all $n\geq15$.
This can be done inductively.
If $3^n>n^6$ then $3\cdot3^{n}>3\cdot n^6$ which is greater than $(n+1)^6$, for $n>15$ (because $3>(1+\frac1n)^6$ for all $n>15$).

Therefore, eventually $0<\frac{n^5}{3^n}<\frac1n$. It follows that $\frac{n^5}{3^n}\xrightarrow[n\to +\infty]{}0$.

Here's an "elementary" proof that doesn't use any theorems such as l'Hôpital. Start off by writing $\frac{n^5}{3^n} = \frac{n^5}{2^n} \left(\frac{2}{3}\right)^n$. The term $\left(\frac{2}{3}\right)^n$ tends to zero as $n \rightarrow \infty$, so we would be done if we showed that the term $\frac{n^5}{2^n}$ is bounded as $n \rightarrow \infty$. It is clear that this term is always positive. The derivative of the function given by $f(n) = \frac{n^5}{2^n}$ is equal to $f'(n) = -2^{-n} n^4 (n \log(2) - 5)$. For $n > \frac{5}{\log 2}$, we have $f'(n) < 0$, so $f(n)$ is decreasing there. In particular, for all $n > \frac{5}{\log 2}$ we get $0 < f(n) \leq f\left(\frac{5}{\log 2}\right)$ and hence $0 < \frac{n^5}{3^n} \leq f\left(\frac{5}{\log 2}\right) \left(\frac{2}{3}\right)^n$. The right hand side tends to zero as $n \rightarrow \infty$ and we can conclude that $\lim_{n \rightarrow \infty} \frac{n^5}{3^n} = 0$.