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.
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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.
So, what is the negation and what trouble are you having using it?Quote:
Originally Posted by renolovexoxo
Do you mean f(x)= 1/x? If not, what do you mean?
Assuming f(x)= 1/x, suppose the limit did exist and were equal to "a". What would the definition of "limit" say?
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."Quote:
Originally Posted by renolovexoxo
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.Quote:
Originally Posted by renolovexoxo
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).
So I can choose any delta, and show that it satisfies the negation? I could even choose an integer or something like that? Sorry, my teacher is incredibly picky.
Quote:
Originally Posted by renolovexoxo
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.Quote:
Originally Posted by emakarov