# Asymptotic notation for a function

I have a question about asymptotic notation in some proof of a claim which I don't fully understand. In the proof there is a part in which they show that there is a function $f:\Bbb{N} \to [0,1]$ such that $$(1-g^2(n))^n \le f(n) \le (1-g^2(n))^n e^{\frac{ng^3(n)}{2(1-g^2(n))}}$$

where $g(n) = \frac{1}{log(n)}$, and from it they conclude that $f(n) = (1-g^2(n))^{n(1-o(1))}$. What does it mean? and why is it true?

It means that $f(n)$ is equal to $(1-g^2(n))^{n(1-h(x))}$
where $h(x)$ is a function such that $\lim\limits_{g\to 0} = 0$.
This is of course the case since we have $0\leq -nh(x) \leq \log_{1-g^2(n)}(e^{\frac{ng^3(n)}{2(1-g^2(n))}})=\frac{\frac{ng^3(n)}{2(1-g^2(n))}}{\log(1-g^2(n))}=\frac{n}{2(1-1/\log(n)^2)\log(1-1/\log(n)^2)1/log(n)^ 3}$
dividing by $-n$ gives:
$0\geq h(x)\geq - \frac{1}{2(1-1/\log(n)^2)\log(1-1/\log(n)^2)1/log(n)^ 3}$.
Which implies $h(x)$ goes to zero as desired.