Big O: Prove that for all $x \leqslant y$, $n^x \in \mathcal O(n^y)$

Question

For all real numbers, if $x \leqslant y, n^x \in \mathcal O(n^y)$.

This is a homework question, so I'm just looking for a little guidance with this question, and not the answer. I understand how to prove Big O questions using actual functions, but this is something new so I'm unsure. I know that I have to find a c and b that make this true, but I can't figure out how to do the algebra to arrive at that conclusion. Any hints with this?

So far, I've only got the first step:

$n^x \le c*(n^y)$

Where can I go from here?

Answer 1

You may find the limit test helpful here. Let $L = \lim_{n \to \infty} \frac{f(n)}{g(n)}$. $f(n) \in O(g(n))$ if $L < \infty$. If $0 < L < \infty$, then $f(n) \in \Theta(g(n))$, which implies $f(n) \in O(g(n))$. Otherwise, if $L = 0$, then $f(n) \in o(g(n))$, which implies that $f(n) \in O(g(n))$.