How can I prove that Cx will intersect x^2

Question

I want to disprove $ cx \geq x^2 \ \forall \ x $ where c is a real number. (i.e. show that x^2 is not O(x) )

So it seems that I can show that the two must intersect at some point ... if I divide both sides of the inequality by x (assuming a nonzero x, since I just need to disprove it somewhere)

$c \geq x \ \forall \ x $ but if x is a real number and c is a real number then there must be some value x that is less than c.

Is this logic sound?

Answer 1

Set $p(x)=x^2-cx$. Then $p(x)=0$ has two roots $0$ and $c$. For all $x\in (0,c)$ we have $p(x)<0$ that is $x^2<cx$. But out of this interval $P(x)>0$ or $x^2>cx$. Therefore for an arbitrary $c$, both inequalities may occur.

Answer 2

For any $c > 0$, you can show two things:

(1) for small enough $x$, $x^2 < cx$;

(2) for all large enough $x$, $x^2 > cx$.

To make your proof rigorous, you have to make the "small enough" and "large enough" explicit.

This implies that $cx$ intersects $x^2$ at least once.