A question about limit superior for function

[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!

 

[mathman]

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

 

[zzzhhh]

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

I cannot understand your reply, could you please explain in more detail? Thanks.

 

[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 (x_k) of irrational numbers converging to a, limsup f(x_k)=0.

 

[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!

 

[Office_Shredder]

It is well known that limit of function can be converted to limit of sequenc

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