Help me please clear this doubt

[Agm]

Could someone please help me understand how the Continutiy Condition [imath]lim_(x->a)f(x)=f(a)[/imath] implies that f(a) is defined.
Because one can clearly have a function that is discontinuous (image below) and have the Continuty Condition be true and the function will still be discontinuous in a!

Thanks to anyone that will help me, this has been eating me for the past couple days.

 

[Harry_the_cat]

If f(a) is undefined, then how can the limit(x->a)f(x) = f(a) ? How can it be equal to something undefined?

 

[Agm]

Isn't the limit(x->a)f(x)=f(a)? As x approaches a, f(x) approaches f(a). It doesn't "touch" it but it does approach it. This happens both from the "right" and the "left".

If f(a) is undefined, then how can the limit(x->a)f(x) = f(a) ? How can it be equal to something undefined?

 

[BigBeachBanana]

Isn't the limit(x->a)f(x)=f(a)? As x approaches a, f(x) approaches f(a). It doesn't "touch" it but it does approach it. This happens both from the "right" and the "left".

Look at the left and right-hand sides individually.
The LHS [imath]\lim_{x \to a}f(x)[/imath]
The limit can exists without [imath]f(a)[/imath] actually exist.

Now, the RHS based on the graph is [imath]f(a)[/imath]=undefined

Put the LHS and RHS together
[imath]\lim_{x \to a}f(x)=f(a)=\text{undefined}\\ \implies \lim_{x \to a}f(x)=\text{undefined}[/imath], thus it fails the continuity condition.

The continuity condition is saying that 1) the limit approaching a must exist, and 2) the function must be defined at a.
When both are met, we can say that it's continuous at a. If one fails, then it's not continuous.

 

[skeeter]

The definition as I recall ...

[imath]\displaystyle \lim_{x \to a} f(x) = f(a) \iff f(x) \text{ is continuous at } x=a[/imath]

Your attached graph is not continuous because [imath]f(a)[/imath] is not defined.

 

[Agm]

Look at the left and right-hand sides individually.
The LHS [imath]\lim_{x \to a}f(x)[/imath]
The limit can exists without [imath]f(a)[/imath] actually exist.

Now, the RHS based on the graph is [imath]f(a)[/imath]=undefined

Put the LHS and RHS together
[imath]\lim_{x \to a}f(x)=f(a)=\text{undefined}\\ \implies \lim_{x \to a}f(x)=\text{undefined}[/imath], thus it fails the continuity condition.

The continuity condition is saying that 1) the limit approaching a must exist, and 2) the function must be defined at a.
When both are met, we can say that it's continuous at a. If one fails, then it's not continuous.

But since f(a)=undefined and the lim (x->a) f(x)=f(a)=undefined the continuity condition is still satisfied though. My point is that the continuity condition does not seem to imply that f(a) is defined, which beats the purpose of the condition in the first place.

 

[Agm]

The definition as I recall ...

[imath]\displaystyle \lim_{x \to a} f(x) = f(a) \iff f(x) \text{ is continuous at } x=a[/imath]

Your attached graph is not continuous because [imath]f(a)[/imath] is not defined.

I know that the graph is not continuous in a, yet lim(x->a) f(x)=f(a) in that same graph.
Textbooks and professors say that this last condition implies that f(a) is defined but that doesnt look like it to me, see the graph. Thats where my doubt comes from

 

[topsquark]

I know that the graph is not continuous in a, yet lim(x->a) f(x)=f(a) in that same graph.
Textbooks and professors say that this last condition implies that f(a) is defined but that doesnt look like it to me, see the graph. Thats where my doubt comes from

No! f(a) does not exist in that graph. There is a hole at x = a so the function isn't defined there. What is true is that [math]\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)[/math] (The limit from below is the same as the limit from above.) But f(a) is equal to neither of those being undefined.

I understand it looks like it should be f(a), but you have to be clear that f(a) doesn't exist according to the graph.

-Dan

 

[Harry_the_cat]

Isn't the limit(x->a)f(x)=f(a)? As x approaches a, f(x) approaches f(a). It doesn't "touch" it but it does approach it. This happens both from the "right" and the "left".

The limit does exist.
f(a) doesn't exist.
So they can't be equal.

For them to be equal, f(a) must exist.

 

[Agm]

The limit does exist.
f(a) doesn't exist.
So they can't be equal.

For them to be equal, f(a) must exist.

Ok, then what is lim(x->a) f(x)?

 

[Agm]

No! f(a) does not exist in that graph. There is a hole at x = a so the function isn't defined there. What is true is that [math]\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)[/math] (The limit from below is the same as the limit from above.) But f(a) is equal to neither of those being undefined.

I understand it looks like it should be f(a), but you have to be clear that f(a) doesn't exist according to the graph.

-Dan

I think I got it actually! Thank you very much!

 

[Dr.Peterson]

Ok, then what is lim(x->a) f(x)?

The limit is at the location of the "hole", which appears to be 2:



It's f(a) that doesn't exist, because of the hole. But put a solid dot at that location, and f becomes continuous:

 

[Harry_the_cat]

Ok, then what is lim(x->a) f(x)?

On your graph, it looks like it's 2 (if each notch signifies 1). But f(a) does not equal 2, because f(a) does not exist.