If $f_n(x)=O(g(x))$ for all $n=1,2,\dots$, then is $\lim_{n \rightarrow \infty} f_n(x)=O(g(x))$? In other words, is Big O notation closed?
What about little O notation? If $f_n(x)=o(g(x))$ for all $n=1,2,\dots$, then is $\lim_{n \rightarrow \infty} f_n(x)=o(g(x))$?
Not necessarily, in the case your constant depends on $n$, then as @Rylee mentioned, you can have $f_n(x)=nx$, $g(x)=x$and in this case $|\lim f_n(x)|=\infty$ (for $x\neq 0$) so, you can't have $\lim f_n(x)=O(g(x))$.
But if your constant is uniform (independent from $n$) for instance $|f_n(x)|\leq 2|g(x)|$, then certainly passing to the limit, you will have $|\limsup f_n(x)|\leq 2|g(x)|$, i.e. $\limsup f_n(x)=O(g(x))$