Real Analysis proof Using definition that f is defined near p

In summary, the limit of a real-valued function f as x approaches p is L if and only if for every real epsilon greater than 0, there exists a real delta greater than 0 such that for all x in the domain of f with 0 < |x - p| < delta, the absolute value of f(x) minus L is less than epsilon. This is the definition of continuity and does not require a proof.
  • #1
kbrono
16
0
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.
 
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  • #2
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)?
 
  • #3
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.
 
  • #4
It seems to me that the proof of the above is just straight from definition :-\ .
 
  • #5
hmm, I am probably overthinking it then
 
  • #6
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. :-\
 

FAQ: Real Analysis proof Using definition that f is defined near p

What is the definition of "near" in real analysis?

In real analysis, "near" refers to a small neighborhood around a point p where a function f is defined.

What is the significance of using the definition to prove a statement in real analysis?

Using the definition to prove a statement in real analysis allows for a rigorous and precise proof that is based on the fundamental properties of real numbers. This approach ensures that the proof is valid and applicable in all cases.

How does one use the definition to prove that a function is continuous at a point p?

To prove continuity at a point p, one must show that for any small neighborhood around p, there exists a small neighborhood around p such that all points within it are mapped to points within the small neighborhood around f(p). This can be shown using the definition of continuity and the properties of real numbers.

Can the definition be used to prove other properties of functions, such as differentiability or integrability?

Yes, the definition can be used to prove other properties of functions, such as differentiability or integrability. By carefully applying the definition and using the properties of real numbers, one can prove these properties in a rigorous manner.

Is the use of the definition necessary in all real analysis proofs?

In most cases, the use of the definition is necessary in real analysis proofs to ensure a rigorous and valid argument. However, in some cases, simpler proofs may exist that do not require the use of the definition. It ultimately depends on the specific statement being proven.

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