When we have $F(n) = \Omega(H(n))$ and $G(n)=\mathcal{O}(H(n))$.
Can we prove that $G(n)/F(n) = \mathcal{O}(1)$?
I tired to use the definitions of $\mathcal{O}$ and $\Omega$ but all I ended up with were two inequalities that I couldn't use.
Thanks
When we have $F(n) = \Omega(H(n))$ and $G(n)=\mathcal{O}(H(n))$.
Can we prove that $G(n)/F(n) = \mathcal{O}(1)$?
I tired to use the definitions of $\mathcal{O}$ and $\Omega$ but all I ended up with were two inequalities that I couldn't use.
Thanks
What you likely found in the inequalities you mention is that we have $F(n)\geq C_F H(n)$ and $G(n)\leq C_G H(n)$. So $$G(n)/F(n)\leq G(n)/(C_F H(n))\leq C_G H(n)/(C_F H_n)=C_G/C_F$$ Is this the application of the definitional inequalities you hadn't yet seen?