Methods for Determining Points of Discontinuity

Determining the type of a point of discontinuity is an important skill in the theory of continuity. This article introduces several common determination methods.

Definition Method

Apply the definition of a point of discontinuity directly:

Steps

  1. Check whether the function is defined at that point

    • If the function is undefined at that point, the point may be a point of discontinuity
  2. Compute the limit of the function at that point

    • Compute lim⁡x→x0f(x)\lim_{x \to x_0} f(x)
    • If the limit does not exist, the point is a point of discontinuity
  3. Compare the limit with the function value

    • If lim⁡x→x0f(x)≠f(x0)\lim_{x \to x_0} f(x) \neq f(x_0), the point is a point of discontinuity

Example

Determine the type of discontinuity of f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2} at x=2x = 2

Steps:

  1. The function is undefined at x=2x = 2
  2. Compute the limit: lim⁡x→2x2−4x−2=lim⁡x→2(x+2)=4\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} (x + 2) = 4
  3. The limit exists but the function value is undefined
  4. Conclusion: x=2x = 2 is a removable discontinuity

Left-Right Limit Method

For piecewise functions or complicated functions, the left-right limit method is more convenient:

Steps

  1. Compute the left limit and the right limit separately

    • Left limit: lim⁡x→x0−f(x)\lim_{x \to x_0^-} f(x)
    • Right limit: lim⁡x→x0+f(x)\lim_{x \to x_0^+} f(x)
  2. Compare the left limit, the right limit and the function value

    • If left limit = right limit = function value, the function is continuous
    • If left limit = right limit ≠ function value, it is a removable discontinuity
    • If left limit ≠ right limit, it is a jump discontinuity
    • If at least one one-sided limit does not exist, it is a discontinuity of the second kind

Example

Determine the type of discontinuity of f(x)={x2,x<12x,x≥1f(x) = \begin{cases} x^2, & x < 1 \\ 2x, & x \geq 1 \end{cases} at x=1x = 1

Steps:

  1. Left limit: lim⁡x→1−f(x)=lim⁡x→1−x2=1\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1
  2. Right limit: lim⁡x→1+f(x)=lim⁡x→1+2x=2\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 2x = 2
  3. Function value: f(1)=2×1=2f(1) = 2 \times 1 = 2
  4. Left limit ≠ right limit
  5. Conclusion: x=1x = 1 is a jump discontinuity

Graphical Method

Determine the type of a point of discontinuity by observing the graph of the function:

Characteristics

  1. Continuous point

    • The graph passes through the point continuously
    • There is no break or jump
  2. Removable discontinuity

    • The graph has a “hole”
    • The function can be made continuous by filling in the “hole”
  3. Jump discontinuity

    • The graph has a “jump”
    • The function values on the two sides are not equal
  4. Discontinuity of the second kind

    • The graph may have a vertical asymptote
    • The function values tend to infinity

Example

Determine the type of discontinuity of f(x)=1xf(x) = \frac{1}{x} at x=0x = 0

Observing the graph:

  • The graph of the function has a vertical asymptote at x=0x = 0
  • It tends to negative infinity on the left and to positive infinity on the right
  • Conclusion: x=0x = 0 is a discontinuity of the second kind

Flowchart for Determining a Point of Discontinuity

Start
  ↓
Is f(x) defined at x₀?
  ↓ No → Compute the limit
  ↓ Yes
  ↓
Compute lim f(x)
  x→x₀
  ↓
Does the limit exist?
  ↓ No → Compute the left-hand and right-hand limits
  ↓ Yes
  ↓
Is the limit = f(x₀)?
  ↓ No → Removable discontinuity
  ↓ Yes
  ↓
Continuous point

Exercises

Exercise 1

Use the definition method to determine the type of discontinuity of f(x)=x2−9x−3f(x) = \frac{x^2 - 9}{x - 3} at x=3x = 3.

point of discontinuityclassification of discontinuities

Approach: Follow the steps of the definition method.

Steps:

  1. The function is undefined at x=3x = 3
  2. Compute the limit: lim⁡x→3x2−9x−3=lim⁡x→3(x+3)=6\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} (x + 3) = 6
  3. The limit exists but the function value is undefined

Answer: x=3x = 3 is a removable discontinuity.

Exercise 2

Use the left-right limit method to determine the type of discontinuity of f(x)={x,x<0x+1,x≥0f(x) = \begin{cases} x, & x < 0 \\ x + 1, & x \geq 0 \end{cases} at x=0x = 0.

point of discontinuityclassification of discontinuities

Approach: Compute the left and right limits separately and compare them.

Steps:

  1. Left limit: lim⁡x→0−f(x)=0\lim_{x \to 0^-} f(x) = 0
  2. Right limit: lim⁡x→0+f(x)=1\lim_{x \to 0^+} f(x) = 1
  3. Function value: f(0)=1f(0) = 1
  4. Left limit ≠ right limit

Answer: x=0x = 0 is a jump discontinuity.

Exercise 3

Use the graphical method to determine the type of discontinuity of f(x)=1x2f(x) = \frac{1}{x^2} at x=0x = 0.

point of discontinuityclassification of discontinuities

Approach: Observe the graph features of the function.

Steps:

  1. The graph of the function has a vertical asymptote at x=0x = 0
  2. It tends to positive infinity on both sides
  3. This is the typical feature of a discontinuity of the second kind

Answer: x=0x = 0 is a discontinuity of the second kind.


Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
x0x_0Mathematical symbolx-sub-zeroThe point at which the type of the discontinuity is identified
f(x)f(x)Mathematical symbolf of xThe function whose discontinuity is being classified
f(x0)f(x_0)Mathematical symbolf of x-sub-zeroThe value of the function at the point, compared with the limit
lim⁡x→x0f(x)\lim_{x \to x_0} f(x)Mathematical symbollimit of f of xThe two-sided limit whose existence is checked first
lim⁡x→x0−f(x)\lim_{x \to x_0^-} f(x)Mathematical symbolleft-hand limitThe left limit computed by the left-right limit method
lim⁡x→x0+f(x)\lim_{x \to x_0^+} f(x)Mathematical symbolright-hand limitThe right limit computed by the left-right limit method
≠\neqMathematical symbolnot equal toThe relation that separates a removable break from a jump
x2−4x−2\frac{x^2 - 4}{x - 2}Mathematical symbolrational expressionThe example whose limit is 4 at the undefined point x=2x = 2
1x\frac{1}{x}Mathematical symbolone over xThe example whose graph has a vertical asymptote at the origin
1x2\frac{1}{x^2}Mathematical symbolone over x squaredThe example that tends to plus infinity on both sides

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