# Homework Help: Negation of Limit

1. Nov 21, 2005

### 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

2. Nov 21, 2005

### 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)."

3. Nov 21, 2005

### Galileo

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

$$\forall \epsilon>0 \exists \delta>0 : |x-a|<\delta \Rightarrow |f(x)-L|<\epsilon$$

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

4. Nov 21, 2005

### soulflyfgm

is this right?

is this the right negation of the statement above?

$$\exists\epsilon>0 \forall \delta>0 : |x-a|<\delta \wedge\|f(x)-L|\geq\epsilon$$

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

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

Last edited: Nov 22, 2005
5. Nov 23, 2005

### 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:

$$\forall \epsilon>0 \exists \delta>0 : 0<|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon$$

So change $|x-a|<\varepsilon$ to $0<|x-a|<\varepsilon$, 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
$$\forall \epsilon>0 \exists \delta>0 : |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon$$

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.

Last edited: Nov 23, 2005
6. Feb 13, 2009

### mrbean

Is this the right negation of a finite limit?

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

Thanks.