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Calculus Question on Removable Discontinuity

  1. Oct 12, 2014 #1
    upload_2014-10-12_22-29-54.png
    I don't entirely understand the question which is why I am posting it here. Anyways, from what the question is asking;we are trying to find the removable discontinuity. This would be plugging in x=0 into both equation and combining them. When this is applied to the first equation, the answer is 0. For the second one, it is 7. So, when the question asks me to combine them, I really don't know what to do with the values. Please Help! I would really appreciate it.
     
  2. jcsd
  3. Oct 12, 2014 #2

    Dick

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    The first expression doesn't give you 0 at x=0. It's undefined. Follow the suggestion and combine the fractions and simplify!
     
  4. Oct 14, 2014 #3

    HallsofIvy

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    A function, f(x), has a "removable discontinuity" at x= a when [itex]\lim_{x\to a} f(x)[/itex] exists but is NOT equal to f(a). You "remove" the discontinuity by redefining f(a) to be equal to [itex]\lim_{x\to a} f(x)[/itex].

    So for this problem, you need to determine what [tex]\lim_{x\to 0}\frac{4}{x}+ \frac{-x+ 16}{x(x- 4)}[/tex] is and redefine f(0) to be that number. As the problem says, start by getting "common denominators" and adding the two fractions.
     
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