- Thread starter
- Banned
- #1
Hi Poirot,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?
Totally confused this problem thinking that \(O\) represents a function.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.
The proof is obviously wrong since \(O\) does not represent a function but a class of functions. Sorry.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.
I mean that the limit of the quotient is a constantCould 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.
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?I mean that the limit of the quotient is a constant
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)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?
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.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)
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.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.
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 \(x\to1\).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.
According to the standard definition, it's both.As for your claim, I am afraid you are mistaken; sinx=O(1) as x tends to 0, not infinity.
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.I still need to explain the calculation that I have seen that states , for example,
as x->0, lim(O(x^2))=0.