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?

  • $\begingroup$ 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. $\endgroup$ Nov 13, 2020 at 22:48

1 Answer 1


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 $g \leqslant C f$.


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