# A question about limit superior for function

1. Aug 26, 2010

### zzzhhh

It is well known that limit of function can be converted to limit of sequence. I wonder if it still holds for limit superior of function. This problem is formulated as follows: For function $$f:\mathbb R\rightarrow\mathbb R$$ and $$a\in\mathbb R$$, define $${\lim\sup}\limits_{x\to a}f(x)$$ to be $$\inf\limits_{\delta>0}(\sup\limits_{0<|x-a|<\delta}f(x))$$. Can we have $${\lim\sup}\limits_{x\to a}f(x)=c$$ iff $${\lim\sup}\limits_{n\to\infty}f(x_n)=c$$ for any sequence $$<x_n>$$ satisfying 1)$$x_n\in\mathbb R$$, 2)$$x_n\to a$$ and 3)$$x_n\ne a$$. I have no idea how to prove it, can you help me? Thanks!

2. Aug 26, 2010

### mathman

It is not true as you stated it. The limsup (x->a) ≤ c for any sequence and = c for at least one sequence.

3. Aug 31, 2010

### zzzhhh

4. Aug 31, 2010

### mathman

Example f(x)=1 for x rational, f(x)=0 for x irrational. Let a=0, limsup(x->a) f(x)=1. Take any sequence (xk) of irrational numbers converging to a, limsup f(xk)=0.

5. Sep 1, 2010

### zzzhhh

A great example, I got it! Thank you!
My original intention is to try to establish the statement "both lim sup f(x) and lim inf f(x) exist and equal c (possibly $$\pm\infty$$) iff lim f(x) exists and equals c" from analogical statement for sequence. But now this approach is not feasible. I then proved the above statement for functions by definition. Thank you again, mathman!

6. Sep 1, 2010

### Office_Shredder

Staff Emeritus
This is only true for functions that are continuous at the point. So it's not surprising that a limsup styled in the same manner would fail for a function that is everywhere discontinuous