A Limit of a function involving big O notation. Fix $t$, I was wondering am I allowed to compute the following limit?
$$\lim_{n\rightarrow \infty}n \left(\frac{-t^2}{2n} + O(\frac{1}{n}) \right)$$
I'm quite confused about such O-notation.. is this limit equal to $\frac{-t^2}{2} ?$
 A: I don't think so. As my understanding, we say $f(n)\in O(\frac{1}{n})$ means $\exists$ constant $C$, s.t. $|f(n)|\leq C|\frac{1}{n}|$ as $n\rightarrow \infty$. But $\lim_{n\rightarrow \infty}n O(\frac{1}{n})$ may not exist. For example, $\frac{1}{n}\sin(n)\in O(\frac{1}{n})$, but $\lim_{n\rightarrow \infty} \sin(n)$ does not exist.
A: As written, the limit cannot be determined. $O\left(\dfrac1n\right)$ represents a function of $n$ whose absolute value is bounded by $\dfrac Cn$ for some constant $C$. Thus, $n\,O\left(\dfrac1n\right)$ is bounded independent of $n$, but does not necessarily tend to a limit.
However, if the expression used little-o, then
$$
\lim_{n\to\infty}n\left(\frac{-t^2}{2n}+o\left(\frac1n\right)\right)=-\dfrac{t^2}{2}
$$
since by definition of little-o,
$$
\lim_{n\to\infty}n\,o\left(\frac1n\right)=0
$$
A: There is not much we are able to say about this limit.
Used in an expression like that, $O\left(\frac{1}{n}\right)$ expresses some function $f(n)$ bounded above by $\frac{c}{n}$. Hence, $$n \left( \frac{-t^2}{2n} + O\left(\frac{1}{n}\right) \right) = n\left(\frac{-t^2}{2n} + f(n)\right) \leq n\left(\frac{-t^2}{2n} + \frac{c}{n}\right) = \frac{-t^2}{2} + c$$.
So, we can say at least that the limit is bounded above by this value $$ \lim_{n \to \infty} n\left(\frac{-t^2}{2n} + O\left(\frac{1}{n}\right)\right) \leq \frac{-t^2}{2} + c$$.
But without knowing exactly which $f(n) \in O\left(\frac{1}{n}\right)$ we can't give a more precise value.
