I'm having trouble seeing why the following holds. Given the conditions that $N=\binom{n}{2}$ and $m$ is a function of $n$ such that $N-m \to \infty$ as $n \to \infty$, why is it that
$$(1+o(1)) \sqrt{\frac{N}{2 \pi m (N-m)}} \ge \frac{1}{10 \sqrt{m}}$$
I can see that there's already a $\sqrt{m}$ in the denominator, and my guess is that the 10 comes from $2 \pi < 10$, but I'm not sure what happens to the $N/(N-m)$ or the $o(1)$. ?