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

Key Questions

  • How do you find asymptotic discontinuity?

    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.

    Wataru · 1 · Oct 5 2014
  • What is a removable Discontinuity?

    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.

    Wataru · · Sep 20 2014
  • How do you determine removable discontinuity for a function?

    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.

    Wataru · · Sep 25 2014
  • What is jump Discontinuity?

    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.

    Amory W. · 1 · Aug 19 2014

Questions

Limits