# Limsup = Liminf = L implies Lim = L, using neighborhoods

1. Oct 16, 2011

### moxy

1. The problem statement, all variables and given/known data

I am to prove that

$\lim_{x\rightarrow t}{f(x)} = L \Longleftrightarrow \limsup{f(t)} = \liminf{f(t)} = L$

where I is an interval in ℝ, and when $f:I → ℝ$, then $\forall t \in I$ we let $N_t$ denote the set of all neighborhoods of t relative to I, and we define:

$\limsup{f(t)} := \inf_{U \in N_t} \sup{\{f(s) | s \in U\}}$

$\liminf{f(t)} := \sup_{U \in N_t} \inf{\{f(s) | s \in U\}}$

3. The attempt at a solution

I'm working on $\lim_{x\rightarrow t}{f(x)} = L \Longleftarrow \limsup{f(t)} = \liminf{f(t)} = L$ at the moment.

So, I've assumed $\limsup{f(t)} = \liminf{f(t)} = L$.

That is,
$\inf_{U \in N_t} \sup{\{f(s) | s \in U\}} := \inf_{U \in N_t} \sup{\{f(U)\}} = L$ and
$\sup_{U \in N_t} \inf{\{f(s) | s \in U\}} := \sup_{U \in N_t} \inf{\{f(U)\}} = L$

I know that sup(f(U)) ≥ L for any U in Nt, so there exists some U1 in Nt such that sup(f(U1)) - ε > L for any ε > 0.
==> sup(f(U1)) > L + ε for all ε > 0

Also, inf(f(U)) ≤ L for any U in Nt, so clearly inf(f(U1)) ≤ L < L + ε for any ε > 0.

sup(f(U1)) > L + ε
inf(f(U1)) < L + ε

I don't think that last step is actually getting me anywhere...I'm completely stuck. Any suggestions would be greatly appreciated. My prof recently introduced liminf, limsup, and neighborhoods, and the new terminology is really messing with my head.

Last edited: Oct 17, 2011
2. Oct 16, 2011

### moxy

When I look at this problem, I just want to use the theorem that for some sequence (tn), lim (tn) exists if and only if limsup (tn) = liminf (tn), and if we know the value of one of these, we know the value of all of them. But I assume that this can't be the proof, considering the book went to the trouble of defining the limsup and liminf in such special ways. Unless the info given is what I need to construct a sequence (tn) that converges to L.

(I'm very confused...I've been working on this problem for over five hours now.)

Last edited: Oct 17, 2011
3. Oct 17, 2011

### moxy

Maybe I'm going about it the wrong way. Is it possible to show that f is continuous at t, then I can say that limx-->t f(x) = f(t) = L? Then I can use the limsup = liminf => lim exists theorem for functions (instead of sequences, like I hinted at in my last post), and it seems that the if and only if statement that I'm trying to prove would follow immediately.

Last edited: Oct 17, 2011