# Negating definition of limit?

by soulflyfgm
Tags: definition, limit, negating
 P: 28 can some one help me out with problem.. Explain using only the defenition and formal logic, wat would be needed to be done to show that , for a particular function f(x) and real number L, L is not the limit of f(x) as x approaches a. [hint : at least get as far as carefully negating the definition of the limit] that was the problem...i really dont know how to start this problem.i KNow the definition of limit which is lim-->a F(x) = L if for every number E > 0 there is a corresponding number G(gamma) >0 such that |F(x) - L|
 Emeritus Sci Advisor PF Gold P: 16,099 Have you gotten as far as carefully negating the definition of the limit?
 P: 28 im sorry but i dont know how to negate that defenition...i will assume that the negation will be for every e<0 there is a number gama <0 such that |f(x) -L| >e whenever 0> |x -a| > gama ....probably that doesnt make sense..but does wat i can think of....i would really apriciate some help!
 Emeritus Sci Advisor PF Gold P: 16,099 Negating definition of limit? What do you know about negating logical sentences? The reason I ask is because the entire procedure for negating a sentence is entirely mechanical -- you just apply the steps. (They should be in your book -- if you tend to forget things like this, you should learn where in your book they are!) By the way, this looks like homework, so you should have posted in the homework help forums. (I'm moving it over there now)
 P: 28 well the book doesnt have any information on that.. the only defenition about negation is this the negation of P denoted ~P, is the propocition "not p" ~P is true exactly when P is false.. .. i used that defenition....wat do u recomend me to do ? any help will be apriciated it. thank you so much
 P: 28 i found this in another threat however i do not know wat he means by convergent sequences. Is something like when u trying to take the limit at an ASYMPTOTE of a fuction? i know that the limit doesnt not exist( or goes to infinitive i cannot recall) is that wat he means by convergent sequence? Let f:I->R and let c in I. I want to negate the statements: "f has limit L at c" and "f is continuous at c". Are these correct? f does not have limit L at c if there exists e>0 such that for some sequence {x_n} converging to c, |f(x_n)-L|>e for every n. f is not continuous at c if there exists e>0 such that for some sequence {x_n} converging to c, |f(x_n)-f(c)|>e for every n. edit: also, what is the negation of "f has a limit at c"?
 Emeritus Sci Advisor PF Gold P: 16,099 If you have a separate question, a separate thread would probably be better. (Otherwise, everyone will get confused) Are you absolutely sure your book doesn't have any useful formulae like: $$\neg(P \vee Q) \equiv \neg P \wedge \neg Q$$ ? And similarly for the other logical operations (including quantifiers?)
 P: 28 would the negation be something like this? lim-->a F(x) = L IF not for every E>0 there isnt a corresponding number g>0 such that |F(x) - L|
 Sci Advisor HW Helper P: 2,586 A function f with domain D has limit L at a point c in D iff For all e > 0, there exists d > 0 such that for all x in D, 0 < |x-c| < d implies |f(x) - f(c)| < e Is it clear to you that the above is the correct definition? Can you express this sentence symbolically? What kind of textbook are you using, and what kind of course is it? You can try to either formalize the above sentence, find its negation by mechanically applying logical rules, and then reinterpreting this new formal sentence into the semi-English that we generally use when doing math (the definition I gave above, for example, is in this semi-English), or you can look at that definition and just "do the logic in your head" to see what the negation would say. As a starter, the thing above says, "for all e > 0, something is true" so the negation would be, "for some e > 0, that thing is false." Does that make sense to you? Continue with that approach if it makes sense to you.
 Emeritus Sci Advisor PF Gold P: 16,099 I guess what we need to do is to derive that the rules are for negating sentences! I had assumed you were in a course for which you would either be learning these, or expected to already know them, but I guess I was wrong about that. (I suppose I could just tell you the rules, but then you wouldn't understand as well) The thing I want to emphasize is that the technique is entirely straightforward -- at any point, there is exactly one rule that applies, so you apply it. (It is similar to differentiating a complicated expression in this regard -- at each step, there is an evident "right" rule to use, so you use it, and eventually you're done!) I'll work through two of them, and see if you can get the rest in exercises. $$\neg \forall x \in P: Q = \exists x \in P: \neg Q$$ (Q, of course, can be a complicated expression, but we don't care exactly what that expression is for the purposes of this rule! Just like the rule (f+g)' = f' + g' doesn't care about what f and g actually are) Suppose you want to negate the claim: For that all x in the class P, Q. Well, when someone is claiming that something is always true, all you have to do to prove him false is to find a single counterexample. That is: There exists an x in the class P for which not Q. So this latter sentence must be the negation of the former. $$\neg(P \implies Q) = P \wedge ~Q$$ Suppose you want to negate the claim: "If P then Q" Well, to disprove an implication like this, all you have to do is to demonstrate that the hypothesis is correct, but the conclusion is not. In other words, you want to prove: P and not Q Here's what you'd need to know to negate any logical sentence. I've left some blank for you to work out as an exercise. (If you want, you can post your results here, and I can check them!) Rule 1: $\neg \forall x \in P: Q \equiv \exists x \in P: \neg Q$ example: "It is not true that all umbrellas are black" is the same as "There exists an umbrella that is not black" Rule 2: $\neg \exists x \in P: Q \equiv ?$ example: "There does not exist a person that is 10 feet tall" is the same as "?" Rule 3: $\neg (P \implies Q) \equiv P \wedge \neg Q$ example: You would reject "If it's blue, then it tastes good" exactly when you would accept "It's blue and it doesn't taste good". Rule 4: $\neg (P \wedge Q) \equiv ?$ example: You would reject "He's big and he's tall" exactly when you would accept "?". Rule 5: $\neg (P \vee Q) \equiv ?$ example: You would reject "He's big or he's tall (or both)" exactly when you would accept "?". Rule 6: $\neg \neg P \equiv ?$ example: "It's not, not blue" is the same as "?"
 P: 28 Rule 1: example: "It is not true that all umbrellas are black" is the same as "There exists an umbrella that is not black" Rule 2: = Vx E P: ~Q example: "There does not exist a person that is 10 feet tall" is the same as "There exist all people that are not 10 feet tall" Rule 3: example: You would reject "If it's blue, then it tastes good" exactly when you would accept "It's blue and it doesn't taste good". Rule 4: ~PV~Q example: You would reject "He's big and he's tall" exactly when you would accept "He is not big or he is not tall". Rule 5: ~P^ ~Q example: You would reject "He's big or he's tall (or both)" exactly when you would accept "?"He is not big and he is not tall” Rule 6: P example: "It's not, not blue" is the same as "Blue" __________________ Now that i know a little bit more here is another try of the negation the negation of "f has a limit L at c" is "f does not have a limit L at c if for some number epsilon < 0 there isnt a corresponding number gama <0zero such that |f(x) -L| > epsilon whenever 0>|x-c|>gama i hope this make more sense now and its correct..let me know if im right Hurkyl thank you so much!!!
Emeritus
PF Gold
P: 16,099
 "There does not exist a person that is 10 feet tall" is the same as "There exist all people that are not 10 feet tall"
That's not quite phrased right -- you shouldn't have that "There exist" in front.

 if for some number epsilon < 0 there isnt a corresponding number gama <0zero such that |f(x) -L| > epsilon whenever 0>|x-c|>gama
That's not quite right -- I have two comments.

First off, remember that

"It is not true that all umbrellas are black" is the same as "There exists an umbrella that is not black", but not "There exists a non-umbrella that is not black"

Secondly, it might help to apply the steps one at a time, rather than trying to do it all at once. In other words, start with

not for every number E > 0 there is a corresponding number G >0 such that |F(x) - L| <E whenever 0< |x-a|<G

And apply the steps one at a time. E.G. you would take "there is a corresponding number G >0 such that |F(x) - L| <E whenever 0< |x-a|<G" to be your Q, and apply the "not for all" rule.

Finally, it might help you to rewrite your "whenever" phrase as an "if, then" phrase.
 P: 28 here is another try not for every number E > 0 there is a corresponding number G >0 such that |F(x) - L| 0 there isnt corresponding numbers G>0 such that if |F(x) - L| 0 there isnt a corresponding number G>0 such that if |F(x) - L|
 P: 28 so am i right?
 Sci Advisor HW Helper P: 2,586 So you're starting with: not for every number E > 0 there is a corresponding number G >0 such that |F(x) - L| 0 there isnt corresponding numbers G>0 such that if |F(x) - L| 0 there isnt a corresponding number G>0 such that if |F(x) - L| 0 there isnt a corresponding number G>0 such that if |F(x) - L| 0 there isnt a corresponding number G>0 such that |F(x) - L|
 P: 28 A function f with domain D has limit L at a point c in D iff For all e > 0, there exists d > 0 such that for all x in D, 0 < |x-c| < d implies |f(x) - f(c)| < e The negation of this statement would be something like this? A function f with domain D does not have a limit at point C in D iff not for every number E > 0 there is a corresponding number G >0 such that |F(x) - L| 0 there is not a corresponding number G >0 such that |F(x) - L| 0 there is a corresponding number G >0 such if |F(x) - L|
 Sci Advisor HW Helper P: 2,586 What is the difference between the following two: not for every number E > 0 there is a corresponding number G >0 such that |F(x) - L| 0 there is not a corresponding number G >0 such that |F(x) - L| 0, there is a neighbourhood U of x such that for all x' in U and for all f in F, d(f(x), f(x')) < e. Symbolically: $$(\forall x \in X)(\forall \epsilon > 0)(\exists U \in T)(x \in U\ \wedge \ (\forall x' \in U)(\forall f \in F)(d(f(x),\, f(x')) < e))$$ The negation of this sentence in words is: it is not that for all x in X, for all e > 0, there is a neighbourhood U of x such that for all x' in U and for all f in F, d(f(x), f(x')) < e. there exists x in X such that it is not that for all e > 0, there is a neighbourhood U of x such that for all x' in U and for all f in F, d(f(x), f(x')) < e. there exists x in X such that there exists e > 0 such that there does not exist a neighbourhood U of x such that for all x' in U and for all f in F, d(f(x), f(x')) < e. there exists x in X such that there exists e > 0 such that for all neighbourhoods U of x, it is not that for all x' in U and for all f in F, d(f(x), f(x')) < e. there exists x in X such that there exists e > 0 such that for all neighbourhoods U of x, there exists x' in U such that it is not true that for all f in F, d(f(x), f(x')) < e. there exists x in X such that there exists e > 0 such that for all neighbourhoods U of x, there exists x' in U such that there exists f in F such that it is not the case that d(f(x), f(x')) < e. there exists x in X such that there exists e > 0 such that for all neighbourhoods U of x, there exists x' in U such that there exists f in F such that d(f(x), f(x')) > e. Symbolically: $$\neg (\forall x \in X)(\forall \epsilon > 0)(\exists U \in T)(x \in U\ \wedge \ (\forall x' \in U)(\forall f \in F)(d(f(x),\, f(x')) < e))$$ $$(\exists x \in X)\neg (\forall \epsilon > 0)(\exists U \in T)(x \in U\ \wedge \ (\forall x' \in U)(\forall f \in F)(d(f(x),\, f(x')) < e))$$ $$(\exists x \in X)(\exists \epsilon > 0)\neg (\exists U \in T)(x \in U\ \wedge \ (\forall x' \in U)(\forall f \in F)(d(f(x),\, f(x')) < e))$$ $$(\exists x \in X)(\exists \epsilon > 0)(\forall U \in T)\neg (x \in U\ \wedge \ (\forall x' \in U)(\forall f \in F)(d(f(x),\, f(x')) < e))$$ $$(\exists x \in X)(\exists \epsilon > 0)(\forall U \in T)(\neg(x \in U)\ \vee \ \neg (\forall x' \in U)(\forall f \in F)(d(f(x),\, f(x')) < e))$$ $$(\exists x \in X)(\exists \epsilon > 0)(\forall U \in T)(x \not \in U\ \vee \ (\exists x' \in U)\neg (\forall f \in F)(d(f(x),\, f(x')) < e))$$ $$(\exists x \in X)(\exists \epsilon > 0)(\forall U \in T)(x \not \in U\ \vee \ (\exists x' \in U)(\exists f \in F)\neg (d(f(x),\, f(x')) < e))$$ $$(\exists x \in X)(\exists \epsilon > 0)(\forall U \in T)(x \not \in U\ \vee \ (\exists x' \in U)(\exists f \in F)(d(f(x),\, f(x')) \not < e))$$ $$(\exists x \in X)(\exists \epsilon > 0)(\forall U \in T)(x \not \in U\ \vee \ (\exists x' \in U)(\exists f \in F)(d(f(x),\, f(x')) \geq e))$$ and if you like, you can go one step further and say: $$(\exists x \in X)(\exists \epsilon > 0)(\forall U \in T)(x \in (X-U)\ \vee \ (\exists x' \in U)(\exists f \in F)(d(f(x),\, f(x')) \geq e))$$
 P: 28 well the question is Explain using only the defenition and formal logic, wat would be needed to be done to show that , for a particular function f(x) and real number L, L is not the limit of f(x) as x approaches a. [hint : at least get as far as carefully negating the definition of the limit] the definition of limit which is lim-->a F(x) = L if for every number E > 0 there is a corresponding number G(gamma) >0 such that |F(x) - L| 0, there exists d > 0 such that for all x in D, 0 < |x-c| < d implies |f(x) - f(c)| < e was not for every number E > 0 there is a corresponding number G >0 such if |F(x) - L| 0 there is a corresponding number G >0 such that |F(x) - L|

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