how to prove that $(n^k) \in \mathcal{O}(n^l)$

Question

Can someone help me to prove that

$(n^k) \in \mathcal{O}(n^l)$ for

$1 \lt k \le l$

I tried to prove it with L'Hopital Rule and I got $ \lim_{n\to 0} \frac{\infty}{\infty} = 0$, but it is still not totally clear to me how it works.

Answer 1

The definition of $f(n) \in \mathcal{O}(g(n))$ is that there are constants $N \geq 0$ and $c \in [0, \infty)$ such that $f(n) \leq c g(n)$ for all $n \geq N$. Let's just apply the definition.

If $1 \leq k \leq l$ then $n^k \leq n^l$ for all $n \geq 1$, so $N = 1$ and $c = 1$ suffice.