Product property of Big O

Question

Trying to prove: If $f(n)$ and $g(n)$ are both $O(h(n))$, then $f(n)*g(n)$ is $O(h^2(n))$.

Understanding so far : The product of upper bounds of functions gives an upper bound for the product of the functions:

proof: If $g_1(n) \le c_1\ f_1(n)$ for $n > n_1$ and $g_2(n) \le c_2f_2(n)$ for $n > n_2$, then $g_1(n)g_2(n) \le c_1c_2f_1(n)f_2(n)$ for $n > \max(n_1,n_2)$ .

Any ideas??

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Answer 1

The definition of big O notation tells us that $\lim_{n\to\infty} \left|\frac{f(n)}{h(n)}\right|$ is a constant and similarly for $g(n)$. Thus $$\lim_{n\to\infty} \left|\frac{f(n)}{h(n)}\right|\left|\frac{g(n)}{h(n)}\right|=\lim_{n\to\infty} \left|\frac{f(n)g(n)}{h(n)^2}\right|$$ is a constant, which means $O(fg)=O(h^2)$.