# Big O notation compare inequality

This seems very basic but can not figure it out. I have

$$a = C_{1} \delta \\ |b| = \sqrt{B} \delta, \ B \neq 0$$

where $$\delta >0$$

so both can be written as

$$a = O(\delta) \\b = O(\delta)$$

and $$\delta \to 0$$

Why can't I say

$$\exists \ C\geq (>?) 0: \quad |a| \leq C |b|$$.

I could write $$\delta = \frac{|b|}{\sqrt{B}}$$

and $$|a| = | C_{1} \delta| \leq | C_{1} | \delta = |b| \frac{| C_{1} |}{\sqrt{B}}$$

Why is this wrong?

• If $C_1$ and $B$ are constants then you are not wrong. But if you only know that $a=O(\delta)=b$ as $\delta\to 0$ then you don't know that $C$ exists. Nov 13, 2020 at 22:48

Let me bring example for sequences. Consider $$g(n)= \begin{cases}{} \frac{1}{n}, & n=2k\\ 0, & n=2k-1 \end{cases}$$ and $$f(n)= \begin{cases}{} 0, & n=2k\\ \frac{1}{n}, & n=2k-1 \end{cases}$$ then both holds $$f,g \in O\left(\frac{1}{n}\right)$$, but you cannot write $$f \leqslant C g$$ or $$g \leqslant C f$$.