removable and nonremovable discontinuities in exercises 35–60, find the -values (if any) at which is not continuous. which of the discontinuities are removable?


  1. This prompt is about removable discontinuities. See the explanation below.

    What is a removable discontinuity?

    A removable discontinuity is a point in a graph where it is not linked but may be made so by filling in a single point.
    It is also possible to define it as follows:
    A discontinuity is detachable at x=a if the limit limxaf(x) exists and is finite. There are two kinds of removable discontinuities. At x=a, the function is undefined.
    It should be noted that a nonremovable discontinuity is one in which the limit of the function does not exist at a given point, i.e. lim xa f(x) does not exist.

    What is the calculation justifying the above answer?

    Part A: Where F(x) = 6/x
    At x = 0
    f(x) = 6/0 = ∞; Thus,
    At x = 0
    f(x) is not defined. It is correct to state therefore, that f is continuous for all real integers or number save “zero”.
    Hence,  f(x) is continuous at each x ∈ R – {α}

    Part B: Where F(x) = 4/(x-6)

    At x = 6
    F(x) = 4/(6-6)
    = 4/0
    = ∞;
    Thus, f (x) is not defined.
    We can state therefore that F is continuous at x ∈ R – { α}
    Part C: Where F (x)
    F(x) = x² – 9
    For each C∈R,
    F(c) = C² = 9
    Thus, F(x) here is defined and continuous. That is F(x) is continuous at x ∈ R
    Part D: Where F(x) x² – 4x + 4
    With respect to every C ∈ R,
    F(c) = C² – 4c + 4
    In this instance as well, F(c) is defined and continuous.

    Thus, F(x) in this case is continuous for all X ∈ R

    Learn more about removable discontinuity:
    Full Question:
    Removable and nonremovable discontinuities. In exercises 35–60, find the x-values (if any) at which f is not continuous. Which of the discontinuities are removable?
    35: f(x) = 6/x
    36: f(x) = 4/(x-6)
    37: f(x) = x² – 9
    38: f(x) x² – 4x + 4

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