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Removable discontinuity solution

  1. Feb 13, 2014 #1
    Problem statement ImageUploadedByPhysics Forums1392328376.269890.jpg


    Revelant equations

    None
    Attempt at a solution
    I know it is discontinuous if the right hand limit doesn't equals the left hand limit? Is that correct?
    The other criteria are
    If f(c) exists, lim f(x) x--> c exists and lim f(x)=f(c)

    I don't really understand what the other criteria mean? Also how will I tell from the piecewise function if it is removable?

    For the graph question number 24 I know the points of discontinuity are the open holes however how do I know if it's removable?

    Any help would be appreciated!!!
     
  2. jcsd
  3. Feb 13, 2014 #2
    Removable means that there's a hole in the graph, as opposed to a break as you would find in a piecewise graph. Removable discontinuities have a hole at a place, and then they're discontinuous immediately to the left and right. They have the missing value (filled in dot) somewhere above or below the hole.
     
  4. Feb 13, 2014 #3

    eumyang

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    Homework Helper

    For a function to be continuous at a point c, three conditions must be met:
    1) f(c) is defined.
    The left graph in my attachment shows an example of a graph where f(c) is NOT defined.
    2) [itex]lim_{x \rightarrow c} f(x)[/itex] exists.
    The middle graph shows an example of a graph where f(c) is defined, but the limit at x = c does not exist.
    3) [itex]lim_{x \rightarrow c} f(x) = f(c)[/itex]
    The right graph shows an example of a graph where f(c) is defined and the limit at x = c exists, but lim f(x) does not equal f(c).

    Hope this helps.
     

    Attached Files:

  5. Feb 13, 2014 #4

    Oh I understand it now. But how can that be used towards how to do those questions? How would I start?
     
  6. Feb 13, 2014 #5
    Here's a good criteria to use:

    If [itex]lim_{x \rightarrow c}[/itex] exists and [itex]lim_{x \rightarrow c} \neq f(c)[/itex], then there's a hole at [itex]c[/itex].

    If [itex]lim_{x \rightarrow c}[/itex] doesn't exist, but [itex]f(c)[/itex] does exist, then there's a break, as in a piece wise graph that breaks between portions of the domain.
     
  7. Feb 13, 2014 #6

    I understand that thank you but how can I use that to solve the question?
     
  8. Feb 13, 2014 #7
    Just apply those formulas to the graphs where there are holes. For example, in 23, there's a discontinuity at x = 0, and in that case the limit exists (f(x) approaches 0 from both sides), but what the graph approaches doesn't equal what the function equals -- i.e. limit approaching c doesn't equal f(c), so then it's a removable discontinuity.
     
  9. Feb 13, 2014 #8

    Ok thank you how do I do that using a piecewise function without a graph though?
     
  10. Feb 13, 2014 #9

    Is it that if the 2 sided limits are different then it is not removable?
     
  11. Feb 13, 2014 #10
    Yeah, if the left and right limits aren't equal, then it can't be removable.
     
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