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

## 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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