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In Uni, we learned about the Landau-symbol $\mathcal{O}$. But the definition confuses me.

$f=\mathcal{O}(g)$ for $x\rightarrow x_0$ if and only if there exists a constant $c>0$ and a neighbourhood $U(x_0)$, so that for every $x \in U(x_o): |f(x)| \le c\cdot|g(x)|$

Now I was thinking that this counts for almost every function. E.g. let's take $f(x) = x^2$ and $f(x)=x$.

Let's also choose a neighbourhood: $U(x_0) = [-10,10]$ for $x_0=0$. Obviously for any $x\in U(x_0)$, I could say that $|x^2| \le c\cdot|x|$ with $c = 10$. My problem is that for any neighbourhood, I could choose a constant c that is high enough to satisfy the criterion.

Isn't the criterion always satisfied simply if c is large enough?

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We cannot have $c$ such that $|x| \leq c|x^{2}|$ for $x$ in some interval around $0$. This is because the inequality fails whenever $0<|x| <\frac 1 c$.

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  • $\begingroup$ Could you explain why it fails? Isn't $|x|≤c|x^2|$ true for a c big enough? $\endgroup$ – Finn Eggers Mar 29 '19 at 12:07
  • $\begingroup$ @FinnEggers What is required is a $c$ and a $\delta$ such that the inequality holds for every $x$ with $|x| <\delta$. The constant $c$ is not allowed to vary with $x$; it is a constant. So you cannot increase $c$ based on what $x$ you pick. $\endgroup$ – Kavi Rama Murthy Mar 29 '19 at 12:12
  • $\begingroup$ That makes sense $\endgroup$ – Finn Eggers Mar 29 '19 at 12:37
  • $\begingroup$ But can c change with Delta? $\endgroup$ – Finn Eggers Mar 29 '19 at 12:39
  • $\begingroup$ Yes, but whatever $c$ you choose there is always an $x$ for which the inequality is violated. $\endgroup$ – Kavi Rama Murthy Mar 29 '19 at 13:13
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No. Let $f(x)=x, g(x)=x^2$ and $x_0=0.$ Suppose that there is $c>0$ and a neigborhood $U$ of $x_0$ such that $|f(x)| \le c\cdot|g(x)|$ for all $x \in U$. This would imply that

$1 \le c|x|$ for all $x \in U$ with $x \ne 0.$

But this is absurd.

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  • $\begingroup$ Could you explain with this is being implied? $\endgroup$ – Finn Eggers Mar 29 '19 at 12:07

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