# Analysis of the asymptotic notation \$ sqrt n = O (log ^ 2n) \$

I try to determine if $$f (n) = sqrt n$$ is in $$O (g (n))$$, $$Omega (g (n))$$, or $$Theta (g (n))$$ from where $$g (n) = log ^ 2n$$,

The answer says only that $$f (n) = theta (g (n))$$ is right, but why not $$f (n) = O (g (n))$$ right, too?

The formal definition is $$f (n) = O (g (n))$$ means $$c * g (n)$$ is an upper bound on $$f (n)$$, So there is a certain constant c, that $$f (n)$$ is always $$leq$$ $$c * g (n)$$ for big enough $$n$$,

If we take n = 100,000 (I think that's big enough), then

$$sqrt {100,000} leq c * log ^ 2100,000$$
$$sim 316 leq c * 25$$

Here we actually see one $$c$$ such as $$c = 15$$ that will satisfy the inequality. That makes no sense, because then it should be according to this logic $$c$$ that will satisfy any inequality. Do I interpret the definitions incorrectly?