Does a limit exist at a point of discontinuity?

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    Discontinuity
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Discussion Overview

The discussion centers around the concept of limits at points of discontinuity, specifically focusing on removable discontinuities in functions. Participants explore whether a limit exists at such points and the implications for continuity.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that a limit can exist at a point of removable discontinuity, but the function may not be continuous at that point.
  • Others clarify that for a limit to exist at a point, the left-hand limit and right-hand limit must be equal, even if the function value at that point is not defined.
  • A participant questions the assertion that the limit does not exist, prompting others to consider the values that the function approaches from both sides of the discontinuity.
  • There is a reiteration of the conditions required for continuity, emphasizing that all three conditions must be satisfied for a function to be continuous at a point.

Areas of Agreement / Disagreement

Participants express differing views on whether a limit exists at a point of removable discontinuity, with some asserting that it does exist while others question this interpretation. The discussion remains unresolved regarding the implications of these limits for continuity.

Contextual Notes

Participants reference specific function behaviors and limits without reaching a consensus on the interpretation of these behaviors in relation to continuity.

grace77
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Just wanted to clear up something :
If there is a removable discontinuity on a graph(1,2) with both a right and left hand portion and let's say the question was : whether lim at x-> 1f(x) exist

Would it be yes it does exist just that it is undefined or?
 
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A removable discontinuity is one in which ##\lim_{x \rightarrow a^-} f(x) = \lim_{x \rightarrow a^+} f(x) = L##, but ##f(a) \neq L##. So on a graph, you represent the discontinuity as a hole, and the actual value f(a) as a dot. It means that the value ##a## is not in the domain of the original function ##f(x)##, but by declaring that ##f(a) = L##, the function can be made continuous.
 
CAF123 said:
A removable discontinuity is one in which ##\lim_{x \rightarrow a^-} f(x) = \lim_{x \rightarrow a^+} f(x) = L##, but ##f(a) \neq L##. So on a graph, you represent the discontinuity as a hole, and the actual value f(a) as a dot. It means that the value ##a## is not in the domain of the original function ##f(x)##, but by declaring that ##f(a) = L##, the function can be made continuous.
ImageUploadedByPhysics Forums1392930823.848271.jpg


So let's say that it is asking does limf(x) at x-> exist for this graph, is the answer no?
 
Why do you think no? In more simplistic terms, as you approach 1 from the left, what value is f(x) tending to? As you approach 1 from the right, what value of f(x) are you tending to?
 
CAF123 said:
Why do you think no? In more simplistic terms, as you approach 1 from the left, what value is f(x) tending to? As you approach 1 from the right, what value of f(x) are you tending to?

Im just getting mixed up with it : is this statement true? A limit exists at a point of discontinuity but it is not continuous.
 
The text that you quoted, grace77, said "[itex]\lim_{x\to a^-} f(x)= \lim_{x\to a^+} f(x)[/itex]" Do you not understand what that means? For x between 0 and 1, f(x)= 2x so [itex]\lim_{x\to 1-} f(x)= \lim_{x\to 1^-} 2x[/itex]. What is that equal to? For x between 1 and 2, f(x)= -2x+ 4 so [itex]\lim_{x\to 1^+} f(x)= \lim_{x\to 1^+} -2x+ 4[/itex]. What is that equal to? What does that tell you about [itex]\lim_{x\to 1} f(x)[/itex]?
 
grace77 said:
Im just getting mixed up with it : is this statement true? A limit exists at a point of discontinuity but it is not continuous.
The limit exists at the point of discontinuity but the function is not continuous there.

In order that f(x) be continuous at x= a, three things must be true:
1) f(a) exists.
2) [itex]\lim_{x\to a} f(x)[/itex] exists.
3) [itex]\lim_{x\to a} f(x)= f(a)[/itex]
 

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