Product of limsup

Question

Let $f(x)$ be positive and increasing and $g(x)$ satisfy $\limsup_x g(x)=1$.

I want to show $\limsup_x f(x) g(x)=\infty$

Is that true and how do i show it?

I'm thinking that since $f(x)$ is monotone and increasing $\limsup _x f(x)=\lim _n f(a_n)$ for any $\{a_n\}_{n\geq 1}$ where $a_n\to \infty$

Since the second limsup exist I also (think I) know that i can find a sequence such that $\limsup _x f(x)=\lim _n f(a_n)$.

I got the feeling this is the right way to go, but how do I conclude?

Edit: okay, $\lim_x f(x)=\infty$ i forgot that wasn't true from the given. Is the statement true now?

Answer 1

If I didn't missunterstood your question, taking $f(x)=\arctan(x)+\pi$ and $g(x)=1$ should be a counterexample.

Answer 2

Another trivial example if the question is stated correctly: $$f(x)=1+\frac{x^2}{x^2+1}\quad\text{and}\quad g(x)=1.$$ Then $$\limsup_{x\to\infty}f(x)g(x)=\limsup_{x\to\infty}f(x)=2<\infty.$$