Proving the Existence and Non-Existence of Limits: A Basic Proof Guide

[icesalmon]

Homework Statement

Proove that if the limit as x -> c of f(x) exists and the limit as x -> c of [ f(x) + g(x) ] does not exist then the limit as x -> c of g(x) does not exist.

Homework Equations

The one thing that is coming to mind is Cauchy's Epsilon-Delta definition of a Limit; the fact that if lim as x -> c of f(x) = L means that for each epsilon > 0 there exists a delta > 0 such that if 0 < | x - c | < delta then | f(x) - L | < epsilon but I'm not given any actual functions or numbers to deal with. I'm not sure how this works, but I'm trying to get used to some basic proofs at this point.

The Attempt at a Solution

I really have no clue about what to do, again, since I'm not given any numbers or actual functions I'm not sure where, or how rather, to start the problem.

 

[PAllen]

You can do it the way you thought of, and that is an educational way to do it. First, do you know how to specify the negation of this limit definition? That is, formally state what it means to say a limit does not exist. I think doing so should give you a clue. You don't need to know any numbers or the nature of f and g to reason from there. That is, do you know how to negate: for every blah there exists a foo.

 

[icesalmon]

from what I understand, the limit as x -> c of f(x) does not exist, when the limit as x -> c from the left of f(x) is not equal to the limit as x -> c from the right of f(x). I don't know how to specify or negate this using the Cauchy definition. Does this mean that the limit of f(x) + g(x) is not equal as x -> c from the left and right?

 

[lanedance]

as pallen hinted, start with the definition of a limit existing in terms of epsilon and delta, then negate to show what is meant by a limit not existing

 

[JThompson]

Here are a few hints:

$P\implies Q$ is equivalent to $\neg P\vee Q$. So the negation of the statement is $P\wedge\neg Q$.

Negated universal quantifiers become existential, and negated existential quantifiers become universal.

 

[icesalmon]

I can't read that JThompson. As in I don't understand it and I'm not sure that it will help. I will just have to play with Cauchy for a while.

 

[lanedance]

lim as x -> c of f(x) = L means that for each epsilon > 0 there exists a delta > 0 such that if 0 < | x - c | < delta then | f(x) - L | < epsilon

negate the above

 

[JThompson]

$P\implies Q$ was supposed to relate to the "if |x-c|<delta, then |f(x)-L|<epsilon" part of the epsilon-delta definition.

Essentially, this statement is equivalent to $|x-c|\geq\delta\mbox{ or } |f(x)-L|<\epsilon$. So negating this portion of the definition returns $|x-c|<\delta\mbox{ and } |f(x)-L|\geq\epsilon$.

"There exists" is the existential quantifier and "for all" is the universal quantifier. Negating "for all epsilon > 0" returns "there exists an epsilon > 0". Negating "there exists a delta > 0" returns "for all delta > 0".

 

[icesalmon]

negate the above

i've tried moving epsilons and deltas all around, and nothing is working.
I'm not sure what you mean by "negate" here?

 

[Maxter]

I think this is easiest done by contradiction.

Assume the hypothesis. Then assume *not* the conclusion, ie that the limit for g does exist.

Then use the limit for f existing and the limit for g existing (standard epsilon/2 proof) to prove that the limit of the sum exists, which is a contradiction.

 

[TPAINE]

Do what Maxter said and use contradiction. If lime g(x) existed, the limit of the sum would exist. Therefore, the limit of g(x) does not exist.