I am reading the wikipedia article that proves the central limit theorem and had a question about one of the steps they take in the proof.(See the article here, the context is not really important to understand the question)
They prove that as $n$ approaches infinity, the characteristic function $\varphi: \mathbb{R} \to \mathbb{R}_{\geq 0}$ becomes:
$$ \varphi(t) = 1 - \dfrac{t^2}{2n} + o\left(\dfrac{t^2}{n}\right) $$
Where $o$ is the little-o notation. Then they claim that:
$$ (\varphi(t))^n = \left[1-\dfrac{t^2}{2n} + o\left(\dfrac{t^2}{n}\right)\right]^n \to e^{-\dfrac{t^2}{2}}, n \to \infty $$
I know that $\lim_{n\to\infty} (1+t/n)^n = e^t$, but how can they prove the above equation where we have $o(t^2 / n)$ inside the brackets? Also, I had never seen the small-o notation in use before; if that term were, say, $O(t^3/n\sqrt{n})$ with the big-O notation, would the proof still be valid?
Replacing $t$ with $-t^2/2$ in the result you're familiar with gives $\lim_{n\to\infty}\left(1-\frac{t^2}{2n}\right)^n=e^{-t^2/2}$. But for the CLT, we need a slightly different result. In terms of the little-$o$ notation you asked about, $1-\frac{t^2}{2n}+o\left(\frac{t^2}{2n}\right)$ denotes a certain set of functions, or an arbitrary element thereof. What the CLT's proof notes is that, considered as a function of $n$ at fixed $t$, $\varphi$ is a function in $1-\frac{t^2}{2n}+o\left(\frac{t^2}{2n}\right)$, and each such function satisfies $\lim_{n\to\infty}\varphi^n=e^{-t^2/2}$. In fact, we can use little-$o$ notation on both sides, viz.$$\left[1-\frac{t^2}{2n}+o\left(\frac{t^2}{2n}\right)\right]^n=e^{-t^2/2n+o(t^2/2n)}.$$The advantage of each use of $o(t^2/2n)$ over $O(t^3/n^{3/2})$ is that we don't need any assumptions in the CLT about moments beyond the variance.