First Kind Discontinuity

A discontinuity of the first kind is an important type of discontinuity; its characteristic feature is that both the left and the right limit exist.

Definition

A discontinuity of the first kind is divided into two types:

Removable Discontinuity

Definition

Characteristics

• The left limit equals the right limit
• The limit value is not equal to the function value (or the function value is undefined)
• The function can be made continuous at that point by redefining the function value

Examples

Example 1: $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$

Analysis:

• The function is undefined at $x = 1$
• $\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x+1) = 2$
• The limit exists but the function value is undefined
• Therefore $x = 1$ is a removable discontinuity

Example 2: $f(x) = \begin{cases} \frac{\sin x}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ at $x = 0$

Analysis:

• The function is defined at $x = 0$, with $f(0) = 0$
• $\lim_{x \to 0} \frac{\sin x}{x} = 1$
• The limit exists but is not equal to the function value
• Therefore $x = 0$ is a removable discontinuity

How to Handle Them

For a removable discontinuity, the function can be made continuous by redefining the function value:

Redefining the function value:

• Define the value of the function at that point to be the limit value
• This makes the function continuous at that point

Example:

• Original function: $f(x) = \frac{x^2 - 1}{x - 1}$
• Redefined: $f(x) = \begin{cases} \frac{x^2 - 1}{x - 1}, & x \neq 1 \\ 2, & x = 1 \end{cases}$
• After the redefinition, the function is continuous at $x = 1$

Jump Discontinuity

Definition

Characteristics

• The left limit is not equal to the right limit
• The graph of the function has a “jump” at that point
• The function cannot be made continuous there by redefining the function value

Examples

Example 1: $f(x) = \begin{cases} x, & x < 0 \\ x + 1, & x \geq 0 \end{cases}$ at $x = 0$

Analysis:

• Left limit: $\lim_{x \to 0^-} f(x) = 0$
• Right limit: $\lim_{x \to 0^+} f(x) = 1$
• Function value: $f(0) = 1$
• The left limit is not equal to the right limit
• Therefore $x = 0$ is a jump discontinuity

Example 2: $f(x) = \begin{cases} x^2, & x < 1 \\ 2x, & x > 1 \\ 1, & x = 1 \end{cases}$ at $x = 1$

Analysis:

• Left limit: $\lim_{x \to 1^-} f(x) = 1$
• Right limit: $\lim_{x \to 1^+} f(x) = 2$
• Function value: $f(1) = 1$
• The left limit is not equal to the right limit
• Therefore $x = 1$ is a jump discontinuity

How to Handle Them

A jump discontinuity cannot be made continuous by redefining the function value, but one can:

Analyse the size of the jump:

• Compute the size of the jump: $|\lim_{x \to x_0^+} f(x) - \lim_{x \to x_0^-} f(x)|$
• Analyse the effect of the jump on the properties of the function

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Exercises

Exercise 1

Determine the type of discontinuity of $f(x) = \frac{x^2 - 9}{x - 3}$ at $x = 3$.

Exercise 2

Determine the type of discontinuity of $f(x) = \begin{cases} x^2, & x < 0 \\ x + 1, & x \geq 0 \end{cases}$ at $x = 0$.

Exercise 3

Determine the type of discontinuity of $f(x) = \begin{cases} \frac{\sin x}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ at $x = 0$.

Exercise 4

Given $f(x) = \begin{cases} x, & x < 1 \\ 2 - x, & x \geq 1 \end{cases}$, determine the type of discontinuity at $x = 1$.

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Summary

Symbols Used in This Article