Prove or disprove: Justify your answer using the def or limit. Real analysis

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Homework Help Overview

The discussion revolves around a limit theorem in real analysis, specifically addressing the behavior of a function f defined on the reals excluding zero, and its limit as x approaches zero. The original poster attempts to prove that if f has a positive limit L at zero, then there exists a positive m such that f(x) is positive for values of x close to zero.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster presents a proof attempt involving the definition of a limit, exploring the implications of a positive limit and the conditions under which f(x) remains positive. Some participants express agreement with the reasoning presented.

Discussion Status

The discussion appears to be progressing positively, with participants acknowledging the original poster's reasoning and expressing understanding of the concepts involved. There is a sense of constructive feedback being exchanged, although no formal consensus or resolution has been reached.

Contextual Notes

No specific constraints or missing information have been noted in the discussion. The focus remains on the mathematical reasoning and definitions related to limits in real analysis.

Hodgey8806
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Homework Statement


Suppose f: ℝ-{0} → ℝ has a positive limit L at zero. Then there exists m>0 such that if 0<|x|<m, then f(x)>0.

Homework Equations


The definition of the limit of a function at a point is: (already assuming f to be a function and c being a cluster point)
A real number L is said to be a limit of f at c if, given any ε>0, there exists a δ>0 such that if x is an element of the domain and 0<|x-c|< δ, then |f(x)-L|<ε.

The Attempt at a Solution


I'm proving this true as:
Spse the limit as x→0 of f = L (being a positive real number.
This implies that given ε>0, there exists δ>0 such that 0<|x-0|< δ, then |f(x) - L|<ε.
Let ε=L,
By definition of the limit of a function at a point there exiss δ>0 such that for x in ℝ - {0}, 0<|x|<δ and |f(x) - L|<L if and only if 0<f(x)<2L.
Thus taking m = δ, then for 0<|x|<m and f(x)>0.

I'm sure it is sloppy, but it does make sense to me here. Please, constructive criticism is welcome! :)
 
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Makes sense to me.
 
I agree this is actually quite well done.
 
Thank you very much! Finally analysis is coming back around to itself so I can sort of see where things are going easier :)
 

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