# What order of growth does a ratio of Bigh-Ohs have?

Say that $f(n) = \cal O(n^2)$ and $g(n) = \cal O(n)$.

If $h(n)=f(n)/g(n)$, is it true that $h(n) =\cal O(n)$?

Is it mathematically correct to say that $h(n) = \cal O(n^2)/ O(n) = O(n)$? if not, what would be the correct way to show this?

• Hence we see why you should always use $\Theta$ when you mean it. – Raphael Apr 14 '14 at 7:55
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Let $f(n)= n^2$ and $g(n)=1$. Then $f(n) = \cal O(n^2)$ and $g(n) = \cal O(n)$.
However, $h(n)=f(n)/g(n)=n^2/1 = n^2 \neq \cal{O}(n)$.
• For an even more extreme example, take $g(n)=n^{-1000000}\;$! – David Richerby Apr 14 '14 at 7:08