Using the definition of $O$(big O) show that $6n^2$ $\notin O(2n)$

Question

the definition of Big $O$ is..

$$f(n) < c g(n) $$ $$6n^2 < c 2n$$

if $c = 1$ and $n = 10$, then...

$$6(10)^2 < 1 \cdot 2(10) $$
$$600 < 20 $$
the above is false

Did i show this properly? is there another way to prove this?

Answer 1

Guide:

• You have to show that no such $c$ exists.

• Compute the limit, $\lim_{n \to \infty} \frac{6n^2}{2n}$ and show that it is not bounded by a constant.

Answer 2

You did not show it properly. You have to show that no such $c$ exists. That goes like follows: given any $c$ you can find and $n$ such that $$6n^2 > c\times 2n.$$