Negation of Limit

soulflyfgm

so is that a good negation of the defenition of the limit?

A function f with domain D doesnt not have limit L at a point c in D iff
not for every number E > 0 there is a corresponding number G >0 such if |F(x) - L| <E then is not the case 0< |x-a|<G
am i right? wat am i doing wrong?
thx so much. I made a new post that way ppl wont get confuse with the other post.
thank u so much

EnumaElish

I am guessing that the statement you want to negate is something like "for each e there is a d such that X(d) implies Y(e)." So the negation should be "for some e there is not any d such that X(d) implies Y(e)."

Galileo

It's easy if you use quantifiers:
Definition of [itex]\lim_{x\to a}f(x)=L[/itex]:

[tex]\forall \epsilon>0 \exists \delta>0 : |x-a|<\delta \Rightarrow |f(x)-L|<\epsilon[/tex]

To negate this, simply use the rules:
[tex]\neg (\forall x:P) \iff \exists x: \neg P[/tex]
[tex]\neg (\exists x:P) \iff \forall x : \neg P[/tex]

soulflyfgm

is this the right negation of the statement above?

[tex]\exists\epsilon>0 \forall \delta>0 : |x-a|<\delta \wedge\|f(x)-L|\geq\epsilon[/tex]

i am also using this fact
~(P=>Q) = P^~Q

how can i illustrate this negation? would it be a function that is not continous at a point such as f(x) = 1/(x+1)?
thank you very much for al ur help!

Galileo

I always seem to forget that f(a) isn't important in the definition.
The function doesn't even have to be defined at the limit point. The correct definition is:

[tex]\forall \epsilon>0 \exists \delta>0 : 0<|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon[/tex]

So change [itex]|x-a|<\varepsilon[/itex] to [itex]0<|x-a|<\varepsilon[/itex], then it's correct.

This is different from continuity! A function is continuous at a if [itex]\lim \limits_{x\to a}f(x)=f(a)[/itex], which says 3 things:
1. The limit exists
2. f(a) is defined
3. The 2 are equal.

More precisely, a function is continuous at x=a if
[tex]\forall \epsilon>0 \exists \delta>0 : |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon[/tex]

The function 1/(x+1) is perfectly continuous everwhere on its domain. The point f(-1) is not defined so it's no problem. If you define f(-1)=0, then it's not continuous anymore.

mrbean

Is this the right negation of a finite limit?

[tex]
\neg(\lim_{x\to a}f(x)=L) \iff \exists \epsilon>0 \forall \delta>0\exists x: 0<|x-a|<\delta \Rightarrow |f(x)-L|\geq\epsilon \vee \neg\exists f(x) [/tex]

Thanks.