Given $f(x)=o(1), g(x)=O(h(x)) \text{ as } x\to a $, show that $f(x)g(x)=o(h(x))$

Question

We are given that $f(x)=o(1)$ as $ x\to a $, $g(x)=O(h(x))$ as $ x\to a $. It is required to prove or show that $f(x)g(x)=o(h(x))$ as $ x\to a $

I just multiplied the two functions to get: $$f(x)g(x)=o(1)O(h(x))=\color{red}{o(1)O(1)h(x) = o(1)h(x)=o(h(x))} \text{ as $x\to a$ } $$

My main question is whether or not the part in red is a valid use of the properties of asymptotic expressions.

Answer 1

Recall that by definition as $x\to a$

$$f(x)=o(1)\iff f(x)=1\cdot \omega_1(x) \quad \omega_1(x)\to 0$$

$$g(x)=O(h(x)) \iff g(x)=h(x)\cdot \omega_2(x) \quad \limsup|\omega_2(x)|\in \mathbb R$$

then

$$f(x)g(x)=h(x)\cdot\omega_1(x)\cdot\omega_2(x)=h(x)\cdot \omega_3(x) \quad \omega_3(x)\to 0$$

indeed we have that $|\omega_2(x)|\le M$ and by squeeze theorem

$$|\omega_3(x)|=|\omega_1(x)\cdot\omega_2(x)|\le M|\omega_1(x)\to 0$$

In the same way we can see that

• $O(h(x))=h(x)\cdot O(1)$

• $o(1)\cdot O(1)=o(1)$