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Removable discontinuity

  1. Feb 20, 2014 #1
    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?
     
  2. jcsd
  3. Feb 20, 2014 #2

    CAF123

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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.
     
  4. Feb 20, 2014 #3

    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?
     
  5. Feb 20, 2014 #4

    CAF123

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    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?
     
  6. Feb 20, 2014 #5
    Im just getting mixed up with it : is this statement true? A limit exists at a point of discontinuity but it is not continuous.
     
  7. Feb 20, 2014 #6

    HallsofIvy

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    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]?
     
  8. Feb 20, 2014 #7

    HallsofIvy

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    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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