In other words, how to prove:

For all real constants $a$ and $b$ such that $a > 1$,
$$\lim_{n\to\infty}\frac{n^b}{a^n} = 0$$

I know the definition of limit but I feel that it's not enough to prove this theorem.

We could prove this by induction on integers $k$:  

$$
\lim_{n \to \infty} \frac{n^k}{a^n} = 0.
$$

The case $k = 0$ is straightforward.  I will leave the induction step to you.  To see how this implies the statement for all real $b$, just note that every real number is less than some integer.  In particular, $b \leq \lceil b \rceil$.  Thus,

$$
0 \leq \lim_{n \to \infty} \frac{n^b}{a^n} \leq \lim_{n \to \infty} \frac{n^{\lceil b \rceil}}{a^n} = 0.
$$

The first inequality follows since all the terms are positive.  The last equality follows from the induction we established previously.
