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Removable discontinuity of /x/

  1. Aug 11, 2011 #1
    /x/ = y is non removable discontinuity
    x


    but is
    /x/^2 = y is a removable discontinuity or not? and why?
    x^2
     
  2. jcsd
  3. Aug 11, 2011 #2

    HallsofIvy

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    For real numbers, x not equal to 0,
    [tex]\frac{|x|^2}{x^2}= 1[/tex].
     
  4. Aug 11, 2011 #3
    You can tell from the graphs of these functions whether the discontinuity is removable or not.

    |x|/x is discontinuous for sure at x = 0

    For positive x the graph is the horizontal line y = 1

    For negative x the graph is the horizontal line y = -1

    This causes a 'jump' discontinuity at x = 0 the y value jumps from -1 to 1 at x = 0

    'Jump' discontinuities are not removable because you can not make the graph continuous by filling in a SINGLE point where the problem occurs. In this case at x = 0

    Infinite discontinuities are also non-removable, because you can not make the graph continuous by filling in a SINGLE point.

    |x^2|/x^2 is discontinuous for sure at x = 0

    however graph of |x^2| is exactly the same as graph of x^2 so absolute value symbol is un-necessary since extra branch is not introduced.

    So considering values very close to 0, on the left of 0 |x^2|/x^2 gives 1, on the right of zero |x^2|/x^2 gives 1. This is not a 'jump' discontinuity (or an infinite discontinuity) and the function can be made continuous by filling in a SINGLE point.

    So |x^2|/x^2 has a removable discontinuity at x= 0

    The key is a discontinuity is removable if you can 'fix' discontinuity using a single point.

    Heres a bonus question...1/|x| ... is the discontinuity at x= 0 removable or non removable and why?
     
  5. Aug 11, 2011 #4
    Think about the definition. A discontinuity is said to be "removable" if it's limits converge to something. In some sense, the limit of a function at a point is how we would "reasonably expect" the function to behave at that point. So, if the limit doesn't converge then it must just be that there's no reason to "reasonably expect" the function to play nice there. If the limit converges, though, then we know exactly how it should behave, so we can simply fix it.

    In the first example, the function behaves differently on either side of 0. Does the second one?
     
  6. Aug 12, 2011 #5
    ok thanks a lot
     
  7. Aug 12, 2011 #6


    1/|x| is discontinuity at zero and non removable.... because we couldn't simplify the equation to remove x w/c is 0
    The graph will move closer but never reach x=0
     
    Last edited: Aug 12, 2011
  8. Aug 12, 2011 #7
    Yes, very good. Notice that this discontinuity is different than the discontinuities of |x|/x and |x^2|/x^2

    For 1/|x|, very close to zero, from the right graph goes to positive infinity as x goes to zero, from the left graph goes to positive infinity as x goes to zero. Infinity is not a number, therefore we cannot 'fill it in' using a single point. Even though the function appears to approach the same place from both sides of zero, the place must be a well defined number otherwise it has no limit, as we say. Limits are very important in continuity.

    Bonus question to the bonus question...

    y = 1/x has a non-removable discontinuity at x = 0, how is this non-removable discontinuity different than the non-removable discontinuity of y = 1/|x|

    Hint* Look at both graphs and ask yourself what is happenning to the y values as x approaches 0 from both sides. :smile:

    http://www.wolframalpha.com/input/?i=plot+1/|x|&asynchronous=false&equal=Submit

    http://www.wolframalpha.com/input/?i=plot+1/x&asynchronous=false&equal=Submit
     
    Last edited: Aug 12, 2011
  9. Aug 12, 2011 #8
    I want to ask a question.Is the way I use in determining the kind of Discontinuity is right ?
    First I look where I look where the function is undefined and then then take the limit at this point if it is not an infinite limit and the limit exist then I say the function has a removable discontinuity at this point since if we redefine the function at this point the definition of continuity apply.or sometimes with rational functions I cancel the common factors of numerator and denominator and adding stipulation let it for now x not equal a if the new function(without stipulation of course ) is continuous at this point then the function has a removable discontinuity at a because if we remove the stipulation the function will be continuous at a and .the act of removing the stipulation is the same as redefining the function at a since the limit of the function is the same as the limit of it in lowest terms with x not equal a
     
    Last edited: Aug 12, 2011
  10. Aug 12, 2011 #9
    That all sounds about correct. A rational function before and after you cancel out those factors are not technically the same function. This is because they have a different domain. (The first one wasn't defined where the denominator had a zero.) However, everywhere that those two, different, functions are defined, they're equal to each other. When you cancel, you find a different function that is *almost* exactly the same as the old one, except that it doesn't have the point of discontinuity. So, you've removed it. Therefore, it must've been a removable discontinuity.
     
  11. Aug 12, 2011 #10
    That is my concept.I only add a stipulation to remember that the function is in lowest terms
     
  12. Aug 13, 2011 #11



    In 1/x
    y can be negative if x is negative and goes closer and closer but will never reach the graph y=0 and x=0
    but in 1/|x|
    y will never be a negative no because |x| is always positive the graph goes closer and closer but will never reach to x=0 and y=0
     
    Last edited: Aug 13, 2011
  13. Aug 13, 2011 #12
    Your statements are correct. You are explaining the behavior of the entire graphs. However, when looking at a discontinuity of a single point, it is not a good idea to look at the entire graph. Your interest should be only for what happens at x = 0 because that is where the problem occurs and more subtle analysis is needed for understanding the behavior at x = 0

    The answer i was looking for is 1/|x| goes to positive infinity from both sides of x = 0

    1/x goes to negative infinity on the left of x = 0, positive infinity on the right of x = 0

    The example was meant to show 2 different kind of infinite non-removable discontinuities.:smile:
     
  14. Aug 14, 2011 #13
    Ok. Thanks a lot :smile:
     
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