# Big 'O' order symbols

#### Poirot

##### Banned
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

##### Well-known member
MHB Math Helper
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.

#### Poirot

##### Banned
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

##### Well-known member
MHB Math Scholar
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

##### Well-known member
MHB Math Helper
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.

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.

#### Poirot

##### Banned
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

##### Well-known member
MHB Math Scholar
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?

#### Poirot

##### Banned
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

##### Well-known member
MHB Math Scholar
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.

#### Poirot

##### Banned
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

##### Well-known member
MHB Math Scholar
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

##### Well-known member
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 ...

Kind regards

$\chi$ $\sigma$

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#### Poirot

##### Banned
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

##### Well-known member
MHB Math Scholar
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.