Is $O(\frac{1}{n}) = o(1)$? Sorry about yet another big-Oh notation question, I just found it very confusing.
If $T(n)=\frac{5}{n}$, is it true that $T(n)=O(\frac{1}{n})$ and $T(n) = o(1)$? I think so because (if $h(n)=\frac{1}{n}$) 
$$
\lim_{n \to \infty} \frac{T(n)}{h(n)}=\lim_{n \to \infty} \frac{\frac{5}{n}}{\frac{1}{n}}=5>0 ,
$$
therefore $T(n)=O(h(n))$. 
At the same time (if $h(n)=1$)
$$
\lim_{n \to \infty} \frac{T(n)}{h(n)}=\frac{(\frac{5}{n})}{1}=0,
$$
therefore $T(n)=o(h(n))$.
Thanks!

If $x_n = O(1/n)$, this means there exists $N$ and $C$ such that for all $n > N$, $|x_n| \le C|1/n|$. Hence
$$
\lim_{n \to \infty} \frac{|x_n|}{1} \le \lim_{n \to \infty} \frac{C|1/n|}{1} = 0.
$$
This means if $x_n = O(1/n)$ then $x_n = o(1)$.
Conversely, it is not true though. Saying that $x_n = o(1)$ only means $x_n \to 0$, but there are sequences that go to zero and are not $O(1/n)$ (think of $1/\sqrt{n}$ for instance).
Hope that helps,
