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?