Thank you for your replay, Peter. Of course, it is small o. I will adjust my question.
As for the function, see the first file below. For the limit, I denoted B=f(n,x). Can you please check if really f(n,x)=g(x)+n/3+o(1)?
Thanks again, Peter. You can see well :)
Both numerator and denominator in my limit are O(n2). If the resulting g(x) is correct or not - I can check it easily, don't waste you time.
In O/o symbolic, you are surely more experienced than me. I included my function just to be sure that your response is in line with it.
Dear Viera and Peter,
By the definition of the limit, for a fixed $x$ we have
f(n,x)=g(x)+n/3+o(1), as Peter said.
As I understand
if we write
$f(x,n)= n/3 + O(1)g(x)$ (or with o(1)), we should say wrt for example n.
For me it means that for a fixed $x$ there is $C=C(x)$ such that $|f(x,n)-n/3| \leq C(x) g(x)$ for all $n \neq 1 $.
But, in general we can not say that there is a uniform constant $C$ which works for all $x$ on the set $(1, \infity)$.
Is this correct?
Dear Miodrag, thank you for your contribution.
Right, the convergence is not uniform in x. However, I hope that what I have is good enough for my aim. In another expression, I need to eliminate n, which works if f(n,x)=g(x)+n/3+o(1).
Before posting my question here, I consulted it with my colleagues and studied corresponding texts on internet. Without any convenient response.
If you add any other idea, it will be nice.
Viera
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