Showing that a limit does not equal some value

[Bipolarity]

Suppose I wanted to show that $\lim_{x→c}f(x) ≠ L$ where L and c are real numbers provided in the problem.

One way I could prove the above would be by showing that for some ε>0, there is some x such that for any δ>0, both $0<|x-c|<δ$ and $|f(x)-L|>ε$ are simultaneously satisfied.My question:

Could I also prove the top statement by showing that for some ε>0, there is some x such that for any δ>0, both $0<|x-c|<δ$ and {f(x) is undefined} are simultaneously satisfied.

BiP

 

[chiro]

Hey BiPolarity.

My understanding is that both statements are linked: the epsilons and delta's need to hold as a set of simultaneous constraints.

The reason for this is that what the limit actually does is say that as you reduce the distance from said point to limit point through |x - c| then the distance between the limit and its value get smaller.

So you have a link between the delta and epsilon that keeps this constraint valid and this is what the limit achieves.

If for some reason you didn't get things narrowing down closer and closer then you wouln't get a limit since things would be not getting closer but either staying the same distance or getting farther away as the input (domain) variables got closer together.

 

[Bipolarity]

Hey BiPolarity.

My understanding is that both statements are linked: the epsilons and delta's need to hold as a set of simultaneous constraints.

The reason for this is that what the limit actually does is say that as you reduce the distance from said point to limit point through |x - c| then the distance between the limit and its value get smaller.

So you have a link between the delta and epsilon that keeps this constraint valid and this is what the limit achieves.

If for some reason you didn't get things narrowing down closer and closer then you wouln't get a limit since things would be not getting closer but either staying the same distance or getting farther away as the input (domain) variables got closer together.

Are you saying that proving either statement proves that the limit does not exist?
Because the statement $|f(x)-L|<ε$ is not so easy to take its inverse (its logical opposite).

The statement $|f(x)-L|<ε$ is true if both the following are true:
1) f(x) is real. (in the context of calculus of real-valued functions)
2) $|f(x)-L|<ε$

Thus, if I can show that f(x) is undefined, in which case it is not a number, and thus not even a real number, have I essentially shown that $|f(x)-L|<ε$ is a false statement?

BiP

 

[chiro]

If f(x) is not even a number then what is it?

This doesn't really make sense, since at the end of the day f(x) and L are both numbers or the same structure.

 

[Norwegian]

My question:

Could I also prove the top statement by showing that for some ε>0, there is some x such that for any δ>0, both $0<|x-c|<δ$ and {f(x) is undefined} are simultaneously satisfied.

The answer is no. The only value for x satisfying the above requirement is x=c, and whether or not f(c) is defined is irrelevant for the existence of the limit.

 

[Bipolarity]

If f(x) is not even a number then what is it?

This doesn't really make sense, since at the end of the day f(x) and L are both numbers or the same structure.

There could be a deleted ε-neighborhood of x, say centered around x=c, in which f(x) is not defined at all points in the deleted neighborhood, except possibly at x=c in which we don't care about f(c), though we certainly care about the behavior of f(x) in the neighborhood about that point. Would this count as a proof that the limit of f as x approaches c does not exist?

BiP

 

[haruspex]

The answer is no. The only value for x satisfying the above requirement is x=c, and whether or not f(c) is defined is irrelevant for the existence of the limit.

Quite so. Bipolarity, you should have written:

for some ε>0, for any δ>0, there is some x (which will depend on δ) such that both 0<|x−c|<δ and |f(x)−L|>ε are simultaneously satisfied.​

 

[Bipolarity]

Quite so. Bipolarity, you should have written:

for some ε>0, for any δ>0, there is some x (which will depend on δ) such that both 0<|x−c|<δ and |f(x)−L|>ε are simultaneously satisfied.​

Yes that was what I wrote in my original post.

My question revolves around the fact that the "disprove a limit" is based on the logical inverse of the actual statement of the limit definition.

For the limit of f(x) to equal some value L about the point x=c, {{{for any ε>0 there is always some δ>0 for which 0<|x-c|<δ --> |f(x)-L|<ε. }}}

To prove that the limit of f(x) does not equal L obviously involves proving the logical inverse of the statement above which is wrapped around curly braces.

The inverse of that statement is:

{{{For some ε>0, there is no δ>0 for which 0<|x-c|<δ necessarily implies |f(x)-L|<ε.}}}

Simplifying the logic:

{{{For some ε>0, for every δ>0 there is some x where 0<|x-c|<δ does not imply |f(x)-L|<ε.}}}

Simplifying further:

{{{For some ε>0, for every δ>0 there is some x where 0<|x-c|<δ but |f(x)-L|<ε is false.}}}

Now consider the last clause of that statement, i.e., the statement:
{{|f(x)-L|<ε is false.}} What exactly does this mean and when is it satisfied? Certainly it is satisfied, i.e. true, when {{|f(x)-L|≥ε}}. But it is also true when f(x) is undefined, since f(x) being undefined automatically means {{|f(x)-L|<ε }} cannot be true.

Thus, can we rewrite as:
{{{For some ε>0, for every δ>0 there is some x where 0<|x-c|<δ but (|f(x)-L|≥ε or f(x) is undefined) }}}

Is that not so?

Thanks to all for the help!

BiP

 

[haruspex]

Yes that was what I wrote in my original post.

No, you wrote

for some ε>0, there is some x such that for any δ>0, both 0<|x−c|<δ and |f(x)−L|>ε are simultaneously satisfied.
​

Do you see the difference? In your version, x is chosen before δ. That can't work, as Norwegian pointed out. δ should be chosen first, then an x (the value of which will depend on δ).

 

[Bipolarity]

No, you wrote

for some ε>0, there is some x such that for any δ>0, both 0<|x−c|<δ and |f(x)−L|>ε are simultaneously satisfied.
​

Do you see the difference? In your version, x is chosen before δ. That can't work, as Norwegian pointed out. δ should be chosen first, then an x (the value of which will depend on δ).

Ah ok I see your point, but you could you please address the major concern of my question which I outlined in my last post?

BiP

 

[haruspex]

Ah ok I see your point, but you could you please address the major concern of my question which I outlined in my last post?

As far as I can see, I just did. In your previous post (#8) you laid out the logic by which you inverted the convergence criterion to obtain a criterion for not converging to a specified value. But an error in that logic led you to the wrong formulation. I've provided you with the correct formulation.
Are you now asking where the error was in your logic? Or is there some other concern I'm missing?

 

[Bipolarity]

As far as I can see, I just did. In your previous post (#8) you laid out the logic by which you inverted the convergence criterion to obtain a criterion for not converging to a specified value. But an error in that logic led you to the wrong formulation. I've provided you with the correct formulation.
Are you now asking where the error was in your logic? Or is there some other concern I'm missing?

I think you are completely missing the point of this thread...
my question was specifically about the part |f(x)-L|<ε. I don't know how I could have made that any clearer in any of my posts on this thread. The confusion between x and the δ was just a pedagogical typo on my part which you kindly pointed out, but that fails to address my question which concerns the logical inverse of the statement |f(x)-L|<ε.

My question has nothing to do with the x vs. δ confusion, its just the above portion of the limit definition, i.e. |f(x)-L|<ε that troubles me.

I am asking this question because in one of my homework problems, I was given a function that is defined at a point x=c, but undefined in a deleted neighborhood around it. So how I could prove that limit of f(x) as x approaches c does not equal f(c) ? I certainly can't find any ε >0 for which I can simultaneously show 0<|x-c|<δ and |f(x)-f(c)|≥ε for any δ>0 since f(x) becomes undefined once δ>0 gets small enough. But I can show that f(x) becomes undefined if that were to happen, which technically satisfies the logical inverse of |f(x)-L|<ε. I was wondering if this overall reasoning is valid...

This is the dilemma which motivated the problem; it has nothing to do with your correction... I'm so sorry.

BiP