1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Removable discontinuity solution

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

    Revelant equations

    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


    User Avatar
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Removable discontinuity solution