Negation of Limit: Am I Right? What Am I Doing Wrong?

[soulflyfgm]

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

A function f with domain D doesn't 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 won't 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 \lim_{x\to a}f(x)=L:

\forall \epsilon&gt;0 \exists \delta&gt;0 : |x-a|&lt;\delta \Rightarrow |f(x)-L|&lt;\epsilon

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

 

[soulflyfgm]

is this right?

is this the right negation of the statement above?

\exists\epsilon&gt;0 \forall \delta&gt;0 : |x-a|&lt;\delta \wedge\|f(x)-L|\geq\epsilon

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

how can i illustrate this negation? would it be a function that is not continuous 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:

\forall \epsilon&gt;0 \exists \delta&gt;0 : 0&lt;|x-a|&lt;\delta \Rightarrow |f(x)-L|&lt;\epsilon

So change |x-a|&lt;\varepsilon to 0&lt;|x-a|&lt;\varepsilon, then it's correct.

This is different from continuity! A function is continuous at a if \lim \limits_{x\to a}f(x)=f(a), 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
\forall \epsilon&gt;0 \exists \delta&gt;0 : |x-a|&lt;\delta \Rightarrow |f(x)-f(a)|&lt;\epsilon

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?

<br /> \neg(\lim_{x\to a}f(x)=L) \iff \exists \epsilon&gt;0 \forall \delta&gt;0\exists x: 0&lt;|x-a|&lt;\delta \Rightarrow |f(x)-L|\geq\epsilon \vee \neg\exists f(x)

Thanks.