Classifying Topics of Discontinuity (removable vs. non-removable)

Key Questions

  • If a function #f(x)# has a vertical asymptote at #a#, then it has a asymptotic (infinite) discontinuity at #a#. In order to find asymptotic discontinuities, you would look for vertical asymptotes. Let us look at the following example.

    #f(x)={x+1}/{(x+1)(x-2)}#

    In order to have a vertical asymptote, the function has to display "blowing up" or "blowing down" behaviors. In the case of a rational function like #f(x)# here, it display such behaviors when the denominator becomes zero.

    By setting the denominator equal to zero,

    #(x+1)(x-2)=0 Rightarrow x=-1,2#

    Now, we have a couple of candidates to consider. Let us make sure that there is a vertical asymptote there.

    Is #x=-1# a vertical asymptote?

    #lim_{x to -1}{(x+1)}/{(x+1)(x-2)}#

    by cancelling out #(x+1)#'s,

    #=lim_{x to -1}1/{x-2}=1/{1-2}=-1 ne pminfty#,

    which means that #x=-1# is NOT a vertical asymptote.

    Is #x=2# a vertical asymptote?

    #lim_{x to 2^+}{x+1}/{(x+1)(x-2)}#

    by cancelling out #(x+1)#'s,

    #=lim_{x to 2^+}1/{x-2}=1/0^+=+infty#,

    which means that #x=2# IS a vertical asymptote.

    Hence, #f# has an asymptotic discontinuity at #x=2#.

    I hope that this was helpful.

  • #f(x)# has a removable discontinuity at #x=a# when #lim_{x to a}f(x)# EXISTS; however, #lim_{x to a}f(a) ne f(a)#. A removable discontinuity looks like a single point hole in the graph, so it is "removable" by redefining #f(a)# equal to the limit value to fill in the hole.

  • Recall that a function #f(x)# is continuous at #a# if

    #lim_{x to a}f(x)=f(a)#,

    which can be divided into three conditions:

    C1: #lim_{x to a }f(x)# exists.
    C2: #f(a)# is defined.
    C3: C1 = C2

    A removable discontinuity occurs when C1 is satisfied, but at least one of C2 or C3 is violated. For example, #f(x)={x^2-1}/{x-1}# has a removable discontinuity at #x=1# since

    #lim_{x to 1}{x^2-1}/{x-1} =lim_{x to 1}{(x+1)(x-1)}/{x-1} =lim_{x to 1}(x+1)=2#,

    but #f(1)# is undefined.

  • #lim_(x->a^-)f(x), lim_(x->a^+)f(x)# are finite and #lim_(x->a^-)f(x)!=lim_(x->a^+)f(x)#. So it occurs when the left and right limit at #a# do not match, then we say #f(x)# has a jump discontinuity at #a#.

    This should not be confused with a point discontinuity where:

    #lim_(x->a^-)f(x)=lim_(x->a^+)f(x)#

    which means

    #lim_(x->a)f(x)# exists

    and:

    #lim_(x->a)f(x)!=f(a)#

    It could be the case that #f(a)# is finite or simply DNE.

Questions