Its straightforward to see that $lim_{n\rightarrow \infty}\frac{n^{m-1}}{n^m} = 0$, $m$ is a fixed positive integer. This meas that $n^m$ grows faster that $n^{m-1}$.
Now, let be $\{a_0,a_1,...,a_m\}$ rational numbers, consider the following limit: $lim_{n\rightarrow \infty} a_0+a_1 n+a_2 n^2+...+a_m n^m$. It would seems to me that this goes to $+\infty$ or $-\infty$. Since $a_m n^m$ grows (decreases) faster than any other summand, we just check the sign of $a_m $. How can I prove this? Any suggestion?
$a_0+a_n n+...+a_m n^{m}= n^{m} (a_m +\frac {a_{m-1}} n+...+\frac {a_0} {n^{m}})=n^{m}b_n$ where $b_n \to a_m$ as $n \to \infty$. Hence the limit is $\infty$ if $a_m >0$ and $-\infty$ if $a_m <0$.