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Discontinuity of Rational Functions

A rational function is a quotient of polynomial functions and is discontinuous at the points where its denominator is zero. Understanding the discontinuity of rational functions is essential for mastering the concept of discontinuity.

Basic Properties

A rational function has the following basic properties:

Domain: all real numbers except the points where the denominator is zero
Continuity: continuous on its domain
Points of discontinuity: discontinuous at the points where the denominator is zero
Asymptotes: may have a vertical asymptote and a horizontal asymptote

Definition of Rational Functions

A function of the form $f(x) = \frac{P(x)} {Q(x)}$ is called a rational function, where $P(x)$ and $Q(x)$ are both polynomial functions and $Q(x) \neq 0$ .

Analysis of the Discontinuity of Rational Functions

Types of Point of Discontinuity

The points of discontinuity of rational functions mainly fall into the following types:

Removable discontinuity: the limit exists but the function value is undefined
Jump discontinuity: the left and right limits exist but are not equal
Infinite discontinuity: the limit tends to infinity

Determining Continuity

The continuity of a rational function $f(x) = \frac{P(x)}{Q(x)}$ :

Continuous on its domain: continuous at every point where $Q(x) \neq 0$
Points of discontinuity: discontinuous at every point where $Q(x) = 0$
Removable discontinuities: if $P(x_0) = Q(x_0) = 0$ and $\lim_{x \to x_0} \frac{P(x)}{Q(x)}$ exists, then the function can be made continuous at that point by defining $f(x_0)$

Common Examples of Rational Functions

1. Simple Rational Functions

Example 1: $f(x) = \frac{1}{x}$

Analysis:

Domain: $x \neq 0$
Discontinuous at $x = 0$
Left limit: $\lim_{x \to 0^-} \frac{1}{x} = -\infty$
Right limit: $\lim_{x \to 0^+} \frac{1}{x} = +\infty$
This is a discontinuity of the second kind

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

Analysis:

Domain: $x \neq 1$
Discontinuous at $x = 1$
Can be simplified to $f(x) = x + 1$ (when $x \neq 1$ )
If we define $f(1) = 2$ , then the function is continuous at $x = 1$

2. Complicated Rational Functions

Example 3: $f(x) = \frac{x^2 + 1}{x}$

Analysis:

Domain: $x \neq 0$
Discontinuous at $x = 0$
Left limit: $\lim_{x \to 0^-} \frac{x^2 + 1}{x} = -\infty$
Right limit: $\lim_{x \to 0^+} \frac{x^2 + 1}{x} = +\infty$
This is a discontinuity of the second kind

Example 4: $f(x) = \frac{1}{x^2}$

Analysis:

Domain: $x \neq 0$
Discontinuous at $x = 0$
Left limit: $\lim_{x \to 0^-} \frac{1}{x^2} = +\infty$
Right limit: $\lim_{x \to 0^+} \frac{1}{x^2} = +\infty$
Although the left and right limits are equal, the function is undefined at that point, so it is discontinuous

Exercises

Exercise 1

Determine the continuity of $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$ .

Reference Answer

rational functioncontinuityfunction continuity

Approach: Check whether the function is defined at that point.

Steps:

When $x = 2$ , the denominator $x - 2 = 0$
Therefore the function is undefined at $x = 2$
The function is discontinuous at $x = 2$

Answer: The function is discontinuous at $x = 2$ , because the function is undefined at that point.

Exercise 2

Determine the continuity of $f(x) = \frac{1}{x^2}$ at $x = 0$ .

Reference Answer

rational functioncontinuityfunction continuity

Approach: Check whether the function is defined at that point and compute the limits.

Steps:

The function is undefined at $x = 0$
Left limit: $\lim_{x \to 0^-} \frac{1}{x^2} = +\infty$
Right limit: $\lim_{x \to 0^+} \frac{1}{x^2} = +\infty$
Although the left and right limits are equal, the function is undefined at that point, so it is discontinuous

Answer: The function is discontinuous at $x = 0$ .

Exercise 3

Let $f(x) = \begin{cases} \frac{x^2 - 1}{x - 1}, & x \neq 1 \\ a, & x = 1 \end{cases}$ . Find the value of the constant $a$ that makes $f(x)$ continuous at $x = 1$ .

Reference Answer

rational functioncontinuityfunction continuity

Approach: Use the definition of continuity so that the function value equals the limit value.

Steps:

Compute the limit: $\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x+1) = 2$
By the definition of continuity: $f(1) = \lim_{x \to 1} f(x)$
Therefore $a = 2$

Answer: $a = 2$ .

Exercise 4

Let $f(x) = \frac{x^2 - 1}{x - 1}$ . Find the intervals of continuity of $f(x)$ .

Reference Answer

rational functioncontinuityfunction continuity

Approach: Analyse the domain and the continuity of the rational function.

Steps:

When the denominator is zero: $x - 1 = 0$ , that is, $x = 1$
Domain: $x \neq 1$ , that is, $(-\infty, 1) \cup (1, +\infty)$
A rational function is continuous on its domain
Therefore the interval of continuity is $(-\infty, 1) \cup (1, +\infty)$

Answer: The interval of continuity is $(-\infty, 1) \cup (1, +\infty)$ .

Exercise 5

Determine the continuity of $f(x) = \frac{x^3 + 1}{x^2 - 1}$ at $x = 1$ .

Reference Answer

rational functioncontinuityfunction continuity

Approach: Check whether the function is defined at that point.

Steps:

When $x = 1$ , the denominator $x^2 - 1 = 0$
Therefore the function is undefined at $x = 1$
The function is discontinuous at $x = 1$

Answer: The function is discontinuous at $x = 1$ , because the function is undefined at that point.

Exercise 6

Let $f(x) = \begin{cases} \frac{x^3 - 1}{x - 1}, & x \neq 1 \\ a, & x = 1 \end{cases}$ . Find the value of the constant $a$ that makes $f(x)$ continuous at $x = 1$ .

Reference Answer

rational functioncontinuityfunction continuity

Approach: Use the definition of continuity so that the function value equals the limit value.

Steps:

Compute the limit: $\lim_{x \to 1} \frac{x^3 - 1}{x - 1} = \lim_{x \to 1} \frac{(x-1)(x^2+x+1)}{x-1} = \lim_{x \to 1} (x^2+x+1) = 3$
By the definition of continuity: $f(1) = \lim_{x \to 1} f(x)$
Therefore $a = 3$

Answer: $a = 3$ .

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation/Explanation	Meaning in This Article
$f(x)$	Mathematical symbol	f of x	The rational function whose discontinuities are analysed
$\frac{P(x)}{Q(x)}$	Mathematical symbol	P over Q of x	The quotient of two polynomial functions that defines the function
$P(x)$	Mathematical symbol	P of x	The numerator polynomial of the quotient
$Q(x)$	Mathematical symbol	Q of x	The denominator polynomial whose zeros are the discontinuity points
$Q(x) \neq 0$	Mathematical symbol	Q of x not equal to zero	The condition that keeps a point inside the domain
$x_0$	Mathematical symbol	x-sub-zero	The point where the denominator vanishes and continuity is tested
$\lim_{x \to x_0} \frac{P(x)}{Q(x)}$	Mathematical symbol	limit of P over Q of x	The limit compared with the function value to classify the break
$\cup$	Mathematical symbol	union	Joins the two pieces of the domain around a removed point
$\infty$	Mathematical symbol	infinity	The value a one-sided limit approaches at a pole
$a$	Constant	a	The constant that restores continuity at a removable discontinuity

Piecewise Function

Discontinuity of Trigonometric Functions

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Discontinuity of Rational Functions

Basic Properties

Definition of Rational Functions

Analysis of the Discontinuity of Rational Functions

Types of Point of Discontinuity

Determining Continuity

Common Examples of Rational Functions

1. Simple Rational Functions

2. Complicated Rational Functions

Exercises

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Summary

Symbols Used in This Article

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