Is There a Connection Between Big 'O' Order Symbols and Limits?

[Poirot1]

I have come across a number of examples in my textbook which seem to suggest that
if f(x)-> L as x-> a, then O(f(x))->L as x->a. Can this be proven?

to clarify, I mean O(f(x))=O(f(x)) as x-> a

 

[Sudharaka]

I have come across a number of examples in my textbook which seem to suggest that
if f(x)-> L as x-> a, then O(f(x))->L as x->a. Can this be proven?

Hi Poirot, :)

Yes. This is a direct consequence of the definition of the Big O notation.

Definition: Let \(a\in\Re\) and \(f\) and \(g\) be two functions of \(x\). Then we write,

\[f(x)=O(g(x))\text{ as }x\to a\,\]

if and only if there exist \(\delta,\,M>0\) such that,

\[|f(x)| \le \; M |g(x)|\text{ for }|x - a| < \delta\]

(Reference: Big O notation - Wikipedia, the free encyclopedia)

Now using the above definition we see that,

\[f(x)=O(f(x))\]

Hence if \(f(x)\rightarrow L\) as \(x\rightarrow a\) then,

\[O(f(x))\rightarrow L\]

Kind Regards,
Sudharaka.

 

[Poirot1]

I accept the truth of this but I doubt your reasoning because I would have thought you need to conside arbitrary g(x) =O(f(x), not take a particular case (when g(x)=f(x)).
Also, any proof would need to distinguish between strictly big O and 'little' o since it is false in the latter case. For example, x^2= o(1) as x-> 0 but as x->0, x^2->0 while
1-> 1.

 

[Evgeny.Makarov]

Could you explain what $O(f(x))\rightarrow L$ means? When we write $g(x) = O(f(x))$, the right-hand side at worst is simply a part of a notation and at best is a class of functions.

 

[Sudharaka]

Could you explain what $O(f(x))\rightarrow L$ means? When we write $g(x) = O(f(x))$, the right-hand side at worst is simply a part of a notation and at best is a class of functions.

Totally confused this problem thinking that \(O\) represents a function. (Headbang)

I accept the truth of this but I doubt your reasoning because I would have thought you need to conside arbitrary g(x) =O(f(x), not take a particular case (when g(x)=f(x)).
Also, any proof would need to distinguish between strictly big O and 'little' o since it is false in the latter case. For example, x^2= o(1) as x-> 0 but as x->0, x^2->0 while
1-> 1.

The proof is obviously wrong since \(O\) does not represent a function but a class of functions. Sorry. :)

 

[Poirot1]

Could you explain what $O(f(x))\rightarrow L$ means? When we write $g(x) = O(f(x))$, the right-hand side at worst is simply a part of a notation and at best is a class of functions.

I mean that the limit of the quotient is a constant

 

[Evgeny.Makarov]

I mean that the limit of the quotient is a constant

The quotient of what? And how can the limit of a function not be a constant? Are you talking not about the limit of a function but the limit of a sequence of functions?

 

[Poirot1]

The quotient of what? And how can the limit of a function not be a constant? Are you talking not about the limit of a function but the limit of a sequence of functions?

we say f(x)=O(g(x)) as x-> a if lim{(f(x)/g(x)}= c, where c is not infinity (to answer your second question)

 

[Evgeny.Makarov]

we say f(x)=O(g(x)) as x-> a if lim{(f(x)/g(x)}= c, where c is not infinity (to answer your second question)

First, this is not the standard definition, which can be found in the link to Wikipedia in post #2. For example, \(\sin(x)=O(1)\) as \(x\to\infty\), but \(\lim_{x\to\infty}\sin(x)/1\) does not exist. Second, this does not answer the question what \(O(f(x))\to L\) means.

 

[Poirot1]

First, this is not the standard definition, which can be found in the link to Wikipedia in post #2. For example, \(\sin(x)=O(1)\) as \(x\to\infty\), but \(\lim_{x\to\infty}\sin(x)/1\) does not exist. Second, this does not answer the question what \(O(f(x))\to L\) means.

I am asking, if F(x) tends to L as x to a, then what does some function that is of order f(x) tend to as x tends to a. As for your claim, I am afraid you are mistaken; sinx=O(1) as x tends to 0, not infinity.

 

[Evgeny.Makarov]

I am asking, if F(x) tends to L as x to a, then what does some function that is of order f(x) tend to as x tends to a.

It can tend to any number because "being of order" allows multiplication by any constant factor. For example, if \(f(x)=x\), then \(f(x)\to1\) as \(x\to1\). If \(g(x)=2x\), then \(g(x)=O(f(x))\), but \(g(x)\to2\) as \(x\to1\).

As for your claim, I am afraid you are mistaken; sinx=O(1) as x tends to 0, not infinity.

According to the standard definition, it's both.

 

[chisigma]

This thread shows clearly the unormous confusion that Edmund Landau...

Edmund Landau - Wikipedia, the free encyclopedia

... was able to introduce in the wenstern mathematical thought... it is not a surprise that, when the 'cultural authorities' are like Edmund Landau, sooner or later an Adolf Hitler necessarly appears (Wasntme)... Kind regards $\chi$ $\sigma$

 

[Poirot1]

I still need to explain the calculation that I have seen that states , for example,
as x->0, lim(O(x^2))=0. Now you can see why I hypothesised as I did. I am grateful for the input exposing my ignorance but for now I just want to focus on the above calculation.

 

[Evgeny.Makarov]

I still need to explain the calculation that I have seen that states , for example,
as x->0, lim(O(x^2))=0.

I assume \(O(f(x))\to L\) as \(x\to a\) means that for every function \(g(x)\) such that \(f(x)=O(g(x))\) it is the case that \(g(x)\to L\) as \(x\to a\). Then \(f(x)\to 0\) as \(x\to a\) indeed implies \(O(f(x))\to 0\) as \(x\to a\). This is because for each \(g(x)\in O(f(x))\) there exists a constant \(C\) and a neighborhood of \(a\) where \(|g(x)|\le C|f(x)|\). Since \(f(x)\to 0\), it follows that \(g(x)\to 0\) as well. However, this is true only when the limit of \(f(x)\) is 0.