I am stuck at understanding the little-$o$ notation used in the proof of CLT using Levy's theorem. Specifically, I want to understand why for fixed $n$, $$ 1 - \frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) = \exp\left( -\frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) \right). $$
Being new to the little $o$ notation, I understand the above equality is saying for any $f$ such that $\frac{n}{t^2}f(t) \to 0$ as $t \to 0$, there exists $g$ such that $\frac{n}{t^2}g(t) \to 0$ as $t \to 0$ and $$ 1 - \frac{t^2}{2n} + f(t) = \exp\left( -\frac{t^2}{2n} + g(t) \right). $$ Knowing $\exp(x) = 1 + x + o(x) = 1 + x + h(x)$ for some $h \sim o(x)$, I see $$ 1 - \frac{t^2}{2n} + f(t) = \exp\left( -\frac{t^2}{2n} + f(t) \right) - h\left( -\frac{t^2}{2n}+f(t)) \right) \stackrel{?}{=}\exp\left( -\frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) \right). $$ My question is why is the last equality true? i.e. What is the function $g$ here?
By definition, $f(t) = o(g(t))$ for $t → 0$ if for every $ε > 0$ there is some $ δ > 0$ such that for every $ |t| < δ $, it holds that $ |f(t)| \leq ε |g(t)|$. Equivalently, there is some function $φ(t)$ defined in a neighbourhood of 0 such that $f(t) = φ(t)g(t)$ and $φ(t) → 0 $ as $t → 0$.
To show that $1 - \frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) = \exp\left( -\frac{t^2}{2n} + o\left( \frac{t^2}{n} \right) \right),$ we observe that $\exp(x) = 1+x+o(x) = 1+x + x φ(x)$ for some function $φ$ as above. Thus we have $$ \exp(x + o(x)) = 1 + x + o(x) + (x + o(x))φ(x + o(x)), $$ and it remains to check that $(x + o(x))φ(x + o(x)) = o(x)$ for $x → 0$. Indeed, $|x + o(x)| \leq 2|x|$ for $x$ in some neighbourhood of 0, hence $φ(x + o(x)) → 0 $ as $x → 0$, implying that for any $ε > 0$, $|(x + o(x))φ(x + o(x))| \leq ε|x|$ for all $x$ in some neighbourhood of 0.