Why does $\lim_{ t\to 0} \frac{o(t^2)}{t} = 0$?

Question

Why does $\lim_{ t\to 0} \frac{o(t^2)}{t} = 0$?

$\sqrt t = o(t^2) \implies \lim_{t\to 0} \frac{\sqrt t}{t} = \infty$

Maybe I don't understand completely the little-o notation.

Answer 1

In a right neighbourhood of zero, it is not true that $\sqrt{t}=o(t^2)$.

We say that $f(t)=o(t^2)$ in a right neighbourhood of zero when: $$ \lim_{t\to 0^+}\frac{f(t)}{t^2} = 0.\tag{1}$$ In particular, the previous identity implies that $\frac{f(t)}{t^2}$ is bounded in in a right neighbourhood of zero, say $\left|\frac{f(t)}{t^2}\right|\leq C,$ hence: $$ \left|\frac{f(t)}{t}\right| = |t|\cdot\left|\frac{f(t)}{t^2}\right|\leq C\,|t|\tag{2} $$ implies: $$ \lim_{t\to 0^+}\frac{f(t)}{t} = 0 \tag{3}$$ as wanted.

Answer 2

If $f(t) = o(t^2)$ as $t\to 0$, then $\lim_{t \to 0} f(t)/t^2 = 0$. Thus

$$\lim_{t\to 0} \frac{f(t)}{t} = \lim_{t\to 0} \frac{f(t)}{t^2}\cdot t = \lim_{t\to 0} \frac{f(t)}{t^2}\cdot \lim_{t\to 0} t = 0.$$

Answer 3

1. You may see $f(t)=o(t^2)$ as any function of the form $$f(t)=t^2 \times \epsilon(t)$$ where $\epsilon(t) \to 0$ as $t\to 0$.

2. You may see $f(t)=O(t^2)$ as any function of the form $$f(t)=t^2 \times \epsilon(t)$$ where there exists $C$ such that $|\epsilon(t) |\leq C$ as $t\to 0$.