Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 3 down vote accepted

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

share|cite|improve this answer

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 $$\limsup_{x\to\infty}f(x)g(x)=\limsup_{x\to\infty}f(x)=2<\infty.$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.