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.

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 $\lim_{n \rightarrow \infty} \frac{n^5}{3^n} = 0$.
