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Continuity and Discontinuity in Calculus: Definitions, Theorems, and Applications

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Continuity and Discontinuity of Functions

Types of Discontinuities

In calculus, understanding the behavior of functions at specific points is crucial. Discontinuities are points where a function is not continuous. There are two main types:

  • Removable Discontinuity: Occurs when a function has a 'hole' at a point, but the limit exists. The discontinuity can be 'fixed' by redefining the function at that point.

  • Nonremovable Discontinuity: Includes jump and infinite discontinuities. These cannot be fixed by redefining the function at a single point.

Examples:

  • Jump discontinuity: $f(x)$ has different left and right limits at $x=c$.

  • Infinite discontinuity: $f(x)$ approaches infinity as $x$ approaches $c$.

  • Removable discontinuity: $f(x)$ is undefined at $x=c$, but $\lim_{x \to c} f(x)$ exists.

Definition of Continuity at a Point

A function $f$ is continuous at a point $x=c$ if all the following conditions are met:

  • I. $f(c)$ exists: The function is defined at $x=c$.

  • II. $\lim_{x \to c} f(x)$ exists: The left and right limits as $x$ approaches $c$ are equal and finite.

  • III. $\lim_{x \to c} f(x) = f(c)$: The value of the function at $c$ matches the limit.

Formula:

  • $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$

Discontinuity at a Point

If any of the above conditions fail, the function is discontinuous at $x=c$. The graph of $f$ will show a break, jump, or hole at $c$.

  • Type 1: $f(c)$ is not defined.

  • Type 2: $\lim_{x \to c} f(x)$ does not exist.

  • Type 3: $\lim_{x \to c} f(x) \neq f(c)$.

Example: Using the Continuity Test

To determine continuity at $x=2$ for $f(x)$:

  • Check if $f(2)$ exists.

  • Compute $\lim_{x \to 2^-} f(x)$ and $\lim_{x \to 2^+} f(x)$.

  • If both limits exist and are equal to $f(2)$, the function is continuous at $x=2$.

Example Calculation:

  • If $f(x) = x^2$ for $x < 2$, $f(x) = 2x$ for $x \geq 2$, then $\lim_{x \to 2^-} f(x) = 4$, $\lim_{x \to 2^+} f(x) = 4$, $f(2) = 4$ → continuous at $x=2$.

Determining Values for Continuity

Sometimes, a parameter must be chosen so that a piecewise function is continuous everywhere.

  • Set $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$ and solve for the parameter.

  • Example: For $f(x) = \begin{cases} x^2 + c, & x < 2 \\ 2x + 5, & x \geq 2 \end{cases}$, set $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)$ to solve for $c$.

Calculation:

  • $\lim_{x \to 2^-} f(x) = 4 + c$, $\lim_{x \to 2^+} f(x) = 9$ → $4 + c = 9$ → $c = 5$

Theorem: Properties of Continuous Functions

If $f$ and $g$ are continuous at $x=c$, then the following combinations are also continuous at $x=c$:

Operation

Result

Sum

$f+g$

Difference

$f-g$

Constant multiples

$kf$ (for any number $k$)

Product

$fg$

Quotient

$f/g$, provided $g(c) \neq 0$

Roots

$\sqrt[n]{f}$, if $n$ is a positive integer and $f$ is defined

Practice: Identifying Discontinuities on a Graph

Given a graph, identify:

  • The $x$-values where discontinuities occur

  • Which step of the continuity test is broken

  • The type of discontinuity (removable, jump, infinite)

Value

The three steps of continuity

Type of Discontinuity

$x=2$

Step 2 broken: limit does not exist

Infinite discontinuity

$x=5$

Step 3 broken: limit does not equal $f(5)$

Jump discontinuity

$x=8$

Step 1 broken: $f(8)$ not defined

Removable discontinuity

Intermediate Value Theorem (IVT)

Statement of IVT

The Intermediate Value Theorem states: If $f$ is continuous on $[a, b]$ and $k$ is any value between $f(a)$ and $f(b)$, then there exists at least one $c$ in $[a, b]$ such that $f(c) = k$.

  • Application: Used to show that a function has a root (zero) in an interval.

Example:

  • Show $f(x) = x^2 - 2 - \cos x$ has a zero in $[0, \pi]$:

  • $f(0) = -1$, $f(\pi) = \pi^2 - 2 + 1 = \pi^2 - 1$

  • Since $f(0) < 0$ and $f(\pi) > 0$, IVT guarantees a $c$ in $[0, \pi]$ such that $f(c) = 0$.

Using IVT to Solve for $c$

Given $f(x)$ and an interval $[a, b]$, check:

  • Is $f$ continuous on $[a, b]$?

  • Does $k$ lie between $f(a)$ and $f(b)$?

  • If yes, IVT applies and there exists $c$ such that $f(c) = k$.

Example:

  • For $f(x) = \frac{x^2 - x}{x^2 - 5x + 6}$, find $c$ such that $f(c) = 0$ on $[2.5, 4]$.

  • Check continuity and solve $x^2 - x = 0$ → $x=0$ or $x=1$ (must be in interval).

Summary Table: Continuity Test and Discontinuity Types

Step

Test

Discontinuity Type

1

$f(c)$ exists?

Removable

2

$\lim_{x \to c} f(x)$ exists?

Infinite or Jump

3

$\lim_{x \to c} f(x) = f(c)$?

Jump

Additional info:

  • Continuity is a foundational concept for calculus, essential for understanding limits, derivatives, and integrals.

  • Piecewise functions often require careful analysis at the points where the formula changes.

  • The IVT is a key tool for proving the existence of solutions to equations within intervals.

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