Second Kind Discontinuity

A discontinuity of the second kind is another important type of discontinuity; its characteristic feature is that at least one one-sided limit fails to exist.

Definition

Definition of a Second Kind Discontinuity

A discontinuity at which at least one of the left and right limits fails to exist is called a discontinuity of the second kind.

Characteristics

  • At least one one-sided limit fails to exist
  • The function may be unbounded in a neighbourhood of that point
  • The graph of the function may have a vertical asymptote at that point

Common Examples

Example 1: Rational Function

Function: f(x)=1xf(x) = \frac{1}{x} at x=0x = 0

Analysis:

  • The function is undefined at x=0x = 0
  • Left limit: lim⁡x→0−1x=−∞\lim_{x \to 0^-} \frac{1}{x} = -\infty
  • Right limit: lim⁡x→0+1x=+∞\lim_{x \to 0^+} \frac{1}{x} = +\infty
  • Neither the left nor the right limit exists (both tend to infinity)
  • Therefore x=0x = 0 is a discontinuity of the second kind

Graph Features:

  • 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

Example 2: Reciprocal Square Function

Function: f(x)=1x2f(x) = \frac{1}{x^2} at x=0x = 0

Analysis:

  • The function is undefined at x=0x = 0
  • Left limit: lim⁡x→0−1x2=+∞\lim_{x \to 0^-} \frac{1}{x^2} = +\infty
  • Right limit: lim⁡x→0+1x2=+∞\lim_{x \to 0^+} \frac{1}{x^2} = +\infty
  • Neither the left nor the right limit exists (both tend to positive infinity)
  • Therefore x=0x = 0 is a discontinuity of the second kind

Graph Features:

  • The graph of the function has a vertical asymptote at x=0x = 0
  • It tends to positive infinity on both sides

Example 3: Trigonometric Function

Function: f(x)=tan⁡xf(x) = \tan x at x=π2x = \frac{\pi}{2}

Analysis:

  • The function is undefined at x=π2x = \frac{\pi}{2}
  • Left limit: lim⁡x→(π2)−tan⁡x=+∞\lim_{x \to (\frac{\pi}{2})^-} \tan x = +\infty
  • Right limit: lim⁡x→(π2)+tan⁡x=−∞\lim_{x \to (\frac{\pi}{2})^+} \tan x = -\infty
  • Neither one-sided limit exists (they tend to infinity with opposite signs)
  • Therefore x=π2x = \frac{\pi}{2} is a discontinuity of the second kind

Graph Features:

  • The graph of the function has a vertical asymptote at x=π2x = \frac{\pi}{2}
  • It tends to positive infinity on the left and to negative infinity on the right

Example 4: Exponential Function

Function: f(x)=e1xf(x) = e^{\frac{1}{x}} at x=0x = 0

Analysis:

  • The function is undefined at x=0x = 0
  • Left limit: lim⁡x→0−e1x=0\lim_{x \to 0^-} e^{\frac{1}{x}} = 0 (because 1x→−∞\frac{1}{x} \to -\infty)
  • Right limit: lim⁡x→0+e1x=+∞\lim_{x \to 0^+} e^{\frac{1}{x}} = +\infty (because 1x→+∞\frac{1}{x} \to +\infty)
  • The right limit does not exist
  • Therefore x=0x = 0 is a discontinuity of the second kind

Graph Features:

  • It tends to 0 on the left
  • It tends to positive infinity on the right

How to Handle Them

A discontinuity of the second kind usually indicates a severe discontinuity of the function in a neighbourhood of that point:

Analyse the Behaviour of the Function

  • Study the behaviour of the function near that point
  • Analyse whether it affects the overall properties of the function
  • Determine the position of the vertical asymptote

Exercises

Exercise 1

Determine the type of discontinuity of f(x)=1x2f(x) = \frac{1}{x^2} at x=0x = 0.

point of discontinuityclassification of discontinuities

Approach: Compute the left and right limits and decide.

Steps:

  1. The function is undefined at x=0x = 0

  2. Left limit: lim⁡x→0−1x2=+∞\lim_{x \to 0^-} \frac{1}{x^2} = +\infty

  3. Right limit: lim⁡x→0+1x2=+∞\lim_{x \to 0^+} \frac{1}{x^2} = +\infty

  4. Neither the left nor the right limit exists (they are infinite)

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

Exercise 2

Determine the type of discontinuity of f(x)=tan⁡xf(x) = \tan x at x=π2x = \frac{\pi}{2}.

point of discontinuityclassification of discontinuities

Approach: Analyse the definition and the limits of the function at that point.

Steps:

  1. f(x)f(x) is undefined at x=π2x = \frac{\pi}{2}
  2. Left limit: lim⁡x→(π2)−tan⁡x=+∞\lim_{x \to (\frac{\pi}{2})^-} \tan x = +\infty
  3. Right limit: lim⁡x→(π2)+tan⁡x=−∞\lim_{x \to (\frac{\pi}{2})^+} \tan x = -\infty
  4. Neither one-sided limit exists (they tend to infinity with opposite signs)

Answer: x=π2x = \frac{\pi}{2} is a discontinuity of the second kind.

Exercise 3

Determine the type of discontinuity of f(x)=1x−1f(x) = \frac{1}{x-1} at x=1x = 1.

point of discontinuityclassification of discontinuities

Approach: Compute the left and right limits and decide.

Steps:

  1. The function is undefined at x=1x = 1

  2. Left limit: lim⁡x→1−1x−1=−∞\lim_{x \to 1^-} \frac{1}{x-1} = -\infty

  3. Right limit: lim⁡x→1+1x−1=+∞\lim_{x \to 1^+} \frac{1}{x-1} = +\infty

  4. Neither the left nor the right limit exists (they tend to infinity)

Answer: x=1x = 1 is a discontinuity of the second kind.

Exercise 4

Determine the type of discontinuity of f(x)=e1xf(x) = e^{\frac{1}{x}} at x=0x = 0.

point of discontinuityclassification of discontinuities

Approach: Compute the left and right limits and decide.

Steps:

  1. The function is undefined at x=0x = 0

  2. Left limit: lim⁡x→0−e1x=0\lim_{x \to 0^-} e^{\frac{1}{x}} = 0

  3. Right limit: lim⁡x→0+e1x=+∞\lim_{x \to 0^+} e^{\frac{1}{x}} = +\infty

  4. The right limit does not exist (it tends to infinity)

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


Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
π\piGreek letterPi (pi)The ratio of a circle’s circumference to its diameter, used to denote angles (such as π2\frac{\pi}{2})

Found an issue with this content?

Tell us what could be improved.