Real Analysis proof Using definition that f is defined near p

[kbrono]

Let (a, b) be an open interval in R, and p a point of (a, b). Let f be a real-valued function defined on all of (a, b) except possibly at p. We then say that the limit of f as x approaches p is L if and only if, for every real ε > 0 there exists a real δ > 0 such that 0 < | x − p | < δ and x ∈ (a, b) implies | f(x) − L | < εI have absolutely no idea how to go about this proof.

Here's my first part attempt

ep = epsilon
d= delta

Suppose f(x) is defined near p and for every ep>0 there exists a d>0 such that for every x in R with 0<|x-p|<d, |f(x)-L|<ep. Because f(x) is near p that means that there is some d>0 such that 0<|x-p|<d and x is to be near p that means there is some d>0 such that 0<|x-p|<d and x is in the domain of f(x). Since we know that x in the domain of f(x) and 0<|x-p|<d for some d>0 then we have |f(x)-L|<ep. Which is the definition that the lim f(x) as x->p = L.

 

[Dick]

That looks to me like a definition, which doesn't require a proof. Because it's a definition of continuity. The only way it could be considered to require a proof is if you have another definition on continuity. What might that be? Or are you trying apply it to a specific function f(x)? In that case, what's f(x)?

 

[kbrono]

Yes I accidentally put the definition. The proof is

Suppose p,L in R and that f is a function. Show that lim f(x) as x->p = L iff f is defined near p and and for every ep>0 there is a d>0 such that for all x in R with 0<|x-p|<d, |f(x)-L|<ep whenever f(x) is defined.This seems likes its mainly technicalities to show it.

 

[╔(σ_σ)╝]

It seems to me that the proof of the above is just straight from definition :-\ .

 

[kbrono]

hmm, I am probably overthinking it then

 

[╔(σ_σ)╝]

Use the definition. If the delta and epsilons exist then that is precisely the definition of the limit of a function converging to a finite real number.

If limit x->c f(x) exist then by definition the epsilons and delta exist. :-\