Negate the definition of the limit of a function, and use it to prove that for the function
f : (0; 1) --> R where f(x) 1/x, lim x-->0 f(x) does not exist.
I think I have the negation, but I am having trouble using it.
The definition of limit for a function I got in class is this:
The limit of f at a exists if there exists a real number L in R such that for every e>0 there exists d>0 such that for every x in the interval with 0<|x-a|<d then |f(x)-L|<e.
I thought the negation is this:
The limit of f at a does not exist if for all real numbers L in R such that for every e>0 there exists d>0 such that there exists xin the interval with 00<|x-a|<d then |f(x)-L|>e
A couple of remarks. "A real number L in R" is an unnecessary repetition. Also, "if" is used only after "then." In the first statement it is clear that you mean "if 0 < |x - a| < d, then |f(x) - L| < e," but the second one is difficult to parse because of nested "such that."
Your definition of limit is correct. In symbolic form it can be expressed as
∃L ∈ ℝ ∀ε > 0 ∃δ > 0 ∀x 0 < |x - a| < δ ⇒ |f(x) - L| < ε (*)
where ∀ is read as "for all," ∃ is read as "there exists," and "P ⇒ Q" is read as "if P, then Q" or "P implies Q."
The phrase "For x such that P(x) we have Q(x)" is expressed symbolically as ∀x P(x) ⇒ Q(x). Since your version of negation contains "for all real numbers L in R such that for every e>0...," it look like it has the form
∀L ∈ ℝ (∀ε ...) ⇒ ...
which is incorrect. "For all ε" should not be located in the premise of an implication.
The rule to negate a statement is easy. Change all quantifiers into their dual ones and negate the inner, quantifier-free part. The negation of some implication P ⇒ Q is P and not Q. The quantifier-free part of (*) is 0 < |x - a| < δ ⇒ |f(x) - L| < ε, so its negation is 0 < |x - a| < δ and |f(x) - L| ≥ ε. The negation of (*) is
∀L ∈ ℝ ∃ε > 0 ∀δ > 0 ∃x 0 < |x - a| < δ and |f(x) - L| ≥ ε
Now, the intuitive meaning that 1/x has a limit L as x -> 0 is that all sufficiently small x are mapped into an ε-neighborhood of L. The fact that L is not a limit would mean that there are x arbitrarily small to 0 such that 1/x are still sufficiently far from L. For example, you can set ε = 1 and show that inside any δ-neighborhood of 0 there are x such that |1/x - L| ≥ 1. Look at the graph of 1/x and see of this is true.
Okay. Thanks. I get a bit lost doing negations of long statements like this one. When I'm constructing my proof how do I choose my delta? I class what we did was really complicated, and I want to make sure that in my proof I have the right idea.
In the negation, delta is universally quantified. So you don't choose a specific delta, you have to prove ∃x 0 < |x - a| < δ and |f(x) - L| ≥ ε for every delta.
Basically, it is sufficient to show that for any δ, |1/x| gets arbitrarily large for 0 < |x| < δ. Therefore, whatever L is, 1/x cannot be within a 1-neighborhood of L for all 0 < |x| < δ. That is, there are points x with 0 < |x| < δ such that 1/x > L + 1 (if L > 0).
When you are proving something of the form "For all δ, P(δ)", you have no control over δ. You need to prove P(δ) for all possible δ. A proof usually starts with, "Fix an arbitrary δ," about which we can't assume anything; the proof must work for this given δ, whatever it is.