Landau big O notation: What does it mean when $|x|=\mathcal O(\epsilon)$ in the limit $\epsilon\to0$?

Question

I have as a result that $|x(t)|=\mathcal O(\epsilon(t))$, and I know that $\lim_{t\to\infty}\epsilon(t)=0$. Does this now imply, that $\lim_{t\to\infty} x(t)=0$? From the definition of the Big O noration I would say yes, but it would be nice if someone could confirm. Cheers!

Answer 1

It is. One of properties of $x(t) = O(\varepsilon(t))$ as $t \to \infty$ (you don't need absolute value here, it's included in $O$) is that for some $c > 0$ and $t_0$, $|x(t)| < c \cdot\varepsilon(t)$ for all $t > t_0$.

Now for any $q$ we need to show that for some $t_1$ we have $|x(t)| < q$ it $t > t_1$. As $\varepsilon(t) \to 0$, there is $t_2$ such that $\forall t > t_2\colon \varepsilon(t) < \frac{q}{c}$. Take $t_1 = \max(t_0, t_2)$. For any $t > t_1$ we have $|x(t)| < c \cdot \varepsilon(t) < c \cdot \frac{q}{c} = q$.