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Discontinuities

Core Definition

A function f(x)f(x) is discontinuous at a point x=cx = c if it violates any of the three mandatory criteria for mathematical continuity:

  1. f(c)f(c) is defined (i.e., cc is in the domain of ff).
  2. lim⁡x→cf(x)\lim_{x \to c} f(x) exists (as a finite real number LL).
  3. lim⁡x→cf(x)=f(c)\lim_{x \to c} f(x) = f(c). If any condition fails, x=cx = c is classified as a point of discontinuity.

The Three Core Types of Discontinuities

Discontinuities are diagnosed and classified based on the behavioral failure of the limit and the function value:

                          Point of Discontinuity at x = c
                                         |
            +----------------------------+----------------------------+
            |                                                         |
Does the limit lim_{x->c} f(x) exist?                      Do the one-sided limits blow up
            |                                              to +/- infinity?
            +----------------+                                        |
            |                |                                        v
           YES               NO                             <a class="text-accent underline font-semibold" href="/garden/infinite-discontinuity">Infinite Discontinuity</a>
            |                |                              (Vertical Asymptote)
   Is f(c) != lim f(x)?      |                              e.g., f(x) = 1/(x-c)
            |                v
            v       Are one-sided limits
<a class="text-accent underline font-semibold" href="/garden/removable-discontinuity">Removable Discontinuity</a> finite but unequal?
(Hole in the graph)          |
e.g., f(x) = (x^2-1)/(x-1)   +------------> YES: <a class="text-accent underline font-semibold" href="/garden/jump-discontinuity">Jump Discontinuity</a>
                                                 (Step in the graph)
                                                 e.g., f(x) = |x|/x

1. Removable Discontinuity (Hole)

  • Diagnosis: lim⁡x→cf(x)=L\lim_{x \to c} f(x) = L exists as a finite number, but either f(c)f(c) is undefined or f(c)≠Lf(c) \neq L.
  • Geometric Appearance: A smooth continuous curve with an isolated single-point hole (and possibly an isolated point plotted elsewhere at x=cx = c).
  • Algebraic Cause: Arises from shared factors in rational functions that evaluate to the indeterminate form 00\frac{0}{0}.
  • Resolution: "Removable" because redefining f(c)=Lf(c) = L completely repairs the discontinuity.

2. Jump Discontinuity (Step)

  • Diagnosis: The left-hand limit lim⁡x→c−f(x)=L1\lim_{x \to c^-} f(x) = L_1 and right-hand limit lim⁡x→c+f(x)=L2\lim_{x \to c^+} f(x) = L_2 both exist as finite numbers, but L1≠L2L_1 \neq L_2.
  • Geometric Appearance: The curve abruptly breaks and steps vertically by a finite jump distance Δy=∣L2−L1∣\Delta y = |L_2 - L_1|.
  • Algebraic Cause: Piecewise-defined functions with mismatched boundary formulas or absolute value quotients like ∣x−c∣x−c\frac{|x-c|}{x-c}.
  • Resolution: Cannot be repaired by redefining a single point.

3. Infinite Discontinuity (Vertical Asymptote)

  • Diagnosis: One or both one-sided limits grow without bound (lim⁡x→c−f(x)=±∞\lim_{x \to c^-} f(x) = \pm\infty or lim⁡x→c+f(x)=±∞\lim_{x \to c^+} f(x) = \pm\infty).
  • Geometric Appearance: The graph shoots upward or downward towards infinity along a vertical asymptote line x=cx = c.
  • Algebraic Cause: Rational expressions evaluating to k0\frac{k}{0} (where k≠0k \neq 0).

(Note: Functions that oscillate infinitely fast, such as sin⁡(1/x)\sin(1/x) near x=0x = 0, represent an Essential / Wildly Oscillating Discontinuity).


Example Problems

Example Problem 1: Classifying Discontinuities Analytically

Classify the type of discontinuity at x=2x = 2 for each function: (a) f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2}
(b) g(x)=1x−2g(x) = \frac{1}{x - 2}
(c) h(x)={x+1:x<22x−3:x≥2h(x) = \begin{cases} x + 1 & : x < 2 \\ 2x - 3 & : x \ge 2 \end{cases}

Solution
  • (a) f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2}: Direct substitution yields 00\frac{0}{0}. Factoring gives lim⁡x→2(x−2)(x+2)x−2=lim⁡x→2(x+2)=4\lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2}(x+2) = 4. The two-sided limit exists, but f(2)f(2) is undefined.
    Classification: Removable Discontinuity (Hole at (2,4)(2, 4)).
  • (b) g(x)=1x−2g(x) = \frac{1}{x - 2}: Direct substitution yields 10\frac{1}{0}. As x→2−x \to 2^-, g(x)→−∞g(x) \to -\infty; as x→2+x \to 2^+, g(x)→+∞g(x) \to +\infty.
    Classification: Infinite Discontinuity (Vertical asymptote at x=2x = 2).
  • (c) h(x)h(x): Left limit: lim⁡x→2−(x+1)=3\lim_{x \to 2^-} (x+1) = 3. Right limit: lim⁡x→2+(2x−3)=1\lim_{x \to 2^+} (2x-3) = 1. Both limits are finite, but 3≠13 \neq 1.
    Classification: Jump Discontinuity (Jump magnitude Δy=2\Delta y = 2).

Connections & References