I am considering the expression $$ n\cdot e^{\mathcal{O}(n\cdot z_n^2)}$$
where the big-O-Notation is meant for $n\to\infty$; $(z_n)$ is a Zero sequence, i.e.$z_n\to 0$ as $n\to\infty$.
I am wondering what happens with this Expression as $n$ is large, i.e. what is $$ \lim_{n\to\infty} n\cdot e^{\mathcal{O}(n\cdot z_n^2)}. $$
Let $f(n)\in\mathcal{O}(n\cdot z_n^2)$ for $n\to\infty$
by definition, this means that there exists some positive real number $M$ and some $n_0$ such that $$ \lvert f(n)\rvert\leq M\lvert nz_n^2\rvert~\forall n\geq n_0. $$
So I think we have to use this in order to compute $$ \lim_{n\to\infty}n e^{f(n)} $$
The limit could be anything $\geq 0$. Consider $n e^{-\sqrt{n}}$, $n e^{c-\log n}$, and $n e^{\sqrt{n}}$.