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Math Help - Removable Discontinuity Experts Required

  1. #1
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    Removable Discontinuity Experts Required

    I'm not sure if I fully understand the removable discontinuity concept. Can someone check this over for me please?

    f(x)=


    {2cos(x/2), x<0
    {ln(x), for 0<x<1
    {2x^3 - x - 1, for x>1

    I know that the lim of 2cos(x/2) as x-->0- = -2
    I know that the lim of 2x^3 - x - 1 as x-->0+ = -16pi^3 +2pi - 1
    So is it correct to say that the lim of ln(x) as x approaches 0 from the right DNE because ln(x) is undefined when x = 0

    Therefore f(x) is discontinuous at x = 0, because the function is undefined at x = 0 and because the limit does not exist.

    Would it also be correct to say that this function is an example of a removable discontinuity, if we redefine f(x)
    to be

    {2cos(x/2), x<0
    {-2, 0<x<1
    {-2, x>1

    How many changes are you allowed to make to a function f(x) in order to make it into a removable discontinuity?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Several things are incorrect in your thinking on this one

    To have a removable discontinuity at a point the limit must exist at that point

    For example if f(x) = sin(x)/x then f(0) is undefined

    However lim sin(x)/x =1
    x->0

    Therefore if we define f(x) = {sin(x)/x if x is not 0
    {1 if x = 0

    then f(x) is continuous at 0


    1. lim2cos(x/2) = 2 not -2
    x->0-

    2. lim ln(x) = - infinity
    x->0+

    Therefore limf(x) DNE
    x->0

    Therefore the discontiuity is non-removable

    at x= 1

    lim ln(x) = 0 and lim 2x^3 - x - 1 = 0
    x->1 x->1

    therefore define f(x) = 0 at x= 1 to remove the discontiuity.

    Also if we define

    { ln(x), for 0<x<1
    {2x^3 - x - 1, for x>1

    or
    { ln(x), for 0<x<1
    {2x^3 - x - 1, for x>1

    then we will have removed the discontinuity since both ln(x) and

    2x^3 - x -1 are defined at x = 1
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  3. #3
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    Awesome answer!

    Thanks a lot.
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