Continuity and Types of Discontinuities in Calculus

Section 2.6: Continuity

Introduction

This section explores the concept of continuity in functions, focusing on how to determine whether a function is continuous at a point or over an interval. It also covers the different types of discontinuities and introduces the Intermediate Value Theorem, a fundamental result in calculus.

Continuity at a Point

Definition

A function f is continuous at a point a if the following holds:

• f(a) is defined.

• The limit $\lim_{x \to a} f(x)$ exists.

• $\lim_{x \to a} f(x) = f(a)$.

In other words, the value of the function at a matches the value that the function approaches as x approaches a.

Continuity Checklist

1. f(a) is defined (a is in the domain of f).

2. $\lim_{x \to a} f(x)$ exists.

3. $\lim_{x \to a} f(x) = f(a)$.

If any of these conditions fail, the function is not continuous at a.

Types of Discontinuities

Main Types

There are four main types of discontinuities:

Type	Description	Example/Notes
Jump	Left- and right-hand limits exist but are not equal; $\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$	Piecewise functions with different values on each side of a
Removable	Limit exists, but $f(a)$ is not equal to the limit or is undefined	Hole in the graph; can "fix" by redefining $f(a)$
Infinite	At least one of the one-sided limits is infinite; $\lim_{x \to a} f(x) = \pm \infty$	Vertical asymptote
Oscillating	Limit does not exist because the function oscillates between values as $x \to a$	Example: $f(x) = \sin(1/x)$ as $x \to 0$

Examples

• Removable Discontinuity: $f(x) = \frac{x^2 - 25}{x + 5}$ at $x = -5$ is undefined, but the limit exists. This is a removable discontinuity.

• Infinite Discontinuity: $f(x) = \frac{3x^2 - 2}{x - 2}$ at $x = 2$ has a vertical asymptote, so the limit is infinite.

Continuity on an Interval

Definition

A function f is continuous on an interval I if it is continuous at every point in I. For closed intervals, continuity at the endpoints means the function is continuous from the right at the left endpoint and from the left at the right endpoint.

Practice Examples

• Polynomials: Always continuous everywhere.

• Rational Functions: Continuous everywhere except where the denominator is zero.

• Trigonometric Functions: Continuous on their domains.

Piecewise Functions and Continuity

Ensuring Continuity

To ensure a piecewise function is continuous at a point where the formula changes, set the left- and right-hand limits equal at that point and solve for any unknowns.

• For $f(x) = \begin{cases} x, & 0 \leq x < 2 \\ k, & x = 2 \end{cases}$, set $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$ to solve for $k$.

Intermediate Value Theorem (IVT)

Theorem Statement

Theorem: If f is continuous on $[a, b]$ and L is any number between $f(a)$ and $f(b)$, then there exists at least one number c in $(a, b)$ such that $f(c) = L$.

Applications

• Used to prove the existence of roots (zeros) of continuous functions on an interval.

• To apply IVT, check:

  • f is continuous on $[a, b]$

  • L is between $f(a)$ and $f(b)$

Example

• Given $f(x) = 10 - 17x$ on $[0, 1]$, $f(0) = 10$, $f(1) = -7$. Since $0$ is between $10$ and $-7$, IVT guarantees a zero in $[0, 1]$.

Summary Table: Types of Discontinuities

Type	Definition	Graphical Feature
Jump	Left and right limits exist but are not equal	Step or break in the graph
Removable	Limit exists, but $f(a)$ is not equal to the limit or undefined	Hole in the graph
Infinite	Limit is infinite	Vertical asymptote
Oscillating	Function oscillates as $x \to a$	No single value approached

Key Formulas and Concepts

• Continuity at a point: $\lim_{x \to a} f(x) = f(a)$

• Intermediate Value Theorem: If $f$ is continuous on $[a, b]$ and $L$ is between $f(a)$ and $f(b)$, then $\exists c \in (a, b)$ such that $f(c) = L$

Practice Problems

1. Determine if $f(x) = \frac{3x^2 - 2}{x - 2}$ is continuous at $x = 2$.

   • Solution: $f(2)$ is undefined, $\lim_{x \to 2} f(x) = +\infty$. Infinite discontinuity.

2. For $g(x) = \sqrt{x - 4}$ at $x = 1$.

   • Solution: $g(1)$ is not defined. Not continuous.

3. For $h(x) = \frac{x^2 - 25}{x + 5}$ at $x = -5$.

   • Solution: $h(-5)$ is undefined, but the limit exists. Removable discontinuity.

Additional Info

• Polynomials and rational functions are continuous on their domains.

• Trigonometric, exponential, and logarithmic functions are continuous on their domains.

• Continuity is a local property; a function can be continuous at some points and not at others.