Why is $1+O(\frac{(\log n)^2}{n}) = 1-o(1)$?

Question

I'm always surprised by the ease with which some authors use aymptotics. Here's the example that brought this up for me today: $1+O(\frac{(\log n)^2}{n}) = 1-o(1)$. I'm sure there's nothing too deep here, but can someone explain their line of reasoning when they see expressions like this?

Answer 1

One may recall that, as $n \to \infty$, $$ u_n=o(v_n) $$ means $$ \lim_{n \to \infty}\frac{u_n}{v_n}=0. $$ Since $$ \lim_{n \to \infty}\frac{\frac{(\log n)^2}{n}}{1}=\lim_{n \to \infty}\frac{(\log n)^2}{n}=0, $$ then, as $n \to \infty$, $$ O\left(\frac{(\log n)^2}{n}\right)=o(1). $$