# Product of limsup

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?

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Your hypothesis on $g$ says that infinitely often, $g(n) > 1/2$ and your hypothesis on $f$ says that for each $N > 0$ there is an $n$ such that $f(n) > N$. Try to combine these. –  Louis Feb 21 '13 at 12:19
@louis from where do you have $f(n)>N$, i don't get them –  Dominic Michaelis Feb 21 '13 at 12:20
@dom: Good point. If $f$ is increasing but bounded, the statement is wrong. (So the desired conclusion makes sense only when there's a typo in the question.) –  Louis Feb 21 '13 at 12:22
Yes, it's true now. What is $\lim_n \bigl[ f(a_n)g(a_n)\bigr]$? –  David Mitra Feb 21 '13 at 12:51
Well.. If both limits exist it would just be the product of the limits. Is that true still? –  Henrik Feb 21 '13 at 12:57
If I didn't missunterstood your question, taking $f(x)=\arctan(x)+\pi$ and $g(x)=1$ should be a counterexample.
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 $$\limsup_{x\to\infty}f(x)g(x)=\limsup_{x\to\infty}f(x)=2<\infty.$$
yes I guess the question is really about behavior near $+\infty$ so my comment above was irrelevant, and I'll remove it. –  coffeemath Mar 2 '13 at 2:49