Following up How to show $\lim_{n\to \infty}\sqrt{n}^n (1 - (1 - 1/(\sqrt{n}^n))^{2^n})/2^n = 1$? , for what $m= f(n)$ does
$$\lim_{n\to \infty}\frac{\sqrt{m}^n \left(1 - \left(1 - \frac{1}{\sqrt{m}^n}\right)^{2^{m}} \right)}{2^m} = 1 \, ?$$
It seems that if $m=(n\ln{n})/5$ then this holds but for $m=n\ln{n}$ it does not. Using the approximation $\left(1 - \frac{1}{\sqrt{m}^n}\right)^{2^{m}} \approx \left(1 - \frac{2^{m}}{\sqrt{m}^n}\right)$ seems to give the wrong answer here.
More generally, consider two positive sequences $(u_n)$ and $(v_n)$ such that $u_n\to0$ and $v_n\to+\infty$, and define $$ r_n=\frac{1-(1-u_n)^{v_n}}{u_nv_n}. $$ The question is to determine whether $r_n\to1$ or not.
On the one hand, for every $u$ in $(0,1)$ and every $v\geqslant1$, $(1-u)^v\geqslant1-uv$ hence, for every $n$ large enough to ensure that $v_n\geqslant1$, one has $r_n\leqslant1$.
On the other hand, for every $u$ in $(0,1)$ and every $v\geqslant1$, $$1-(1-u)^v=\int_0^uv(1-w)^{v-1}\mathrm dw\geqslant\int_0^uv(1-u)^{v-1}\mathrm dw=vu(1-u)^{v-1},$$ hence for every $n$ large enough to ensure that $v_n\geqslant1$, one has $r_n\geqslant(1-u_n)^{v_n-1}$, in particular, $$ \log r_n\geqslant(v_n-1)\log(1-u_n)\sim-v_nu_n. $$ If $u_nv_n\to0$, this is enough to conclude that $r_n\to1$. (Note that the sign $\sim$ is used above in its strictest, most rigorous, sense.)
The situation here is that $u_n=m(n)^{-n/2}$ with $m(n)\to\infty$ and $v_n=2^{m(n)}$ hence the conditions $u_n\to0$ and $v_n\to+\infty$ always hold and, considering the logarithms, one sees that the condition $u_nv_n\to0$ holds if and only if $$ n\log m(n)-2\log2\cdot m(n)\to+\infty. $$ If $m(n)\sim cn$ for some positive $c$, this holds for every $c$ hence $r_n\to1$.
If $m(n)\sim cn\log n$ for some positive $c$, this holds if and only if $c\leqslant c^*$ with $$ c^*=\frac1{2\log2}\approx0.72. $$ For example, if $c=\frac15$, then $c\leqslant c^*$ hence $r_n\to1$, while, if $c=1$, then $c\gt c^*$ hence the computations above do not allow to conclude (and actually $r_n\to0$).
