Little $o$ notation in the proof of central limit theorem. In the proof im reading for the CLT, they seem to be using the following claim:

If $h(t)=o(t^2)$ then if $$g(n)\stackrel{\text{def}}{=}h\Big(\frac{t}{\sigma\sqrt{n}}\Big)$$ we have that $g(n)=o(\frac{1}{n})$.

More explicitly, I have a function $h$ that satisfies $$\lim_{t \rightarrow \infty} \frac{h(t)}{t^2} = 0$$
(and I seem to know nothing else about $h$) and then they are telling me that
$$\lim_{n \rightarrow \infty} h\Big(\frac{t}{\sigma\sqrt{n}}\Big) n = 0 .$$
But I don't see how I can say anything about $\lim_{x \rightarrow 0} h(x)$ just from knowing $h(t)=o(t^2)$.
Is the claim true? How can I show it?
 A: They likely meant to say $h(t)\in o(t^2)$ as $t\to0$ rather than $t\to\infty$, since otherwise the claim is false: take $h(t)=t$ (so that $h(t)\in o(t^2)$ as $t\to\infty$), then $g(n) = \frac t{\sigma\sqrt{n}}$ is not in $o(\frac1n)$ as $n\to\infty$ since $\lim_{n\to\infty}\frac{tn}{\sigma\sqrt n}=\infty$.
However, if $h(t)\in o(t^2)$ as $t\to0$, then the claim is indeed true since
$$
\lim_{n\to\infty}h\left(\frac t{\sigma\sqrt n}\right)n = \frac{t^2}{\sigma^2}\lim_{s\to0}\frac{h(s)}{s^2} = 0
$$
using the change of variables $s = \frac t{\sigma\sqrt n}$, showing that $g(n)\in o(\frac1n)$ as $n\to\infty$.
A: By definition
$$o(f)=\left\{g:\exists N \in \mathbb{N}, \exists \varepsilon (n), \lim_{n \to \infty}\varepsilon (n)=0, n>N \Rightarrow \ g(n)=\varepsilon (n) f(n)\right\}$$
So $h(t)=o(t^2)$, i.e. $h(t) \in o(t^2)$, assumed you mean case $t \to 0,$ gives $ h\left(\frac{1}{n} \right)=\varepsilon \left(\frac{1}{n} \right) \left(\frac{1}{n} \right)^2$, for $n>N$.
Now we have $g(n) = h\Big(\frac{t}{\sigma\sqrt{n}}\Big) = \varepsilon\Big(\frac{t}{\sigma\sqrt{n}}\Big)\cdot \Big(\frac{t}{\sigma\sqrt{n}}\Big)^2 = \varepsilon\Big(\frac{t}{\sigma\sqrt{n}}\Big)\cdot \Big(\frac{t}{\sigma}\Big)^2 \frac{1}{n}$
