Continuity, Limits, and Types of Discontinuity

Date: September 1, 2026
Observance: National Burnt Ends Day
Today in History:
- Aaron Burr acquitted (1807)
- Wreck of the Titanic found (1985)
Number of the Day: $84$
- Prime Factorization: $84 = 2^2 \times 3 \times 7$
- Property: $84$ is the smallest number with eight representations as a sum of two primes.
Fun Fact: The French language has seventeen different words for 'surrender'.
Quote of the Day: "First the doctor told me the good news: I was going to have a disease named after me." — Steve Martin
Today's Weather: A stray shower or thunderstorm is possible, high $86^\circ\text{F}$ .

Limit Evaluation Problem: Evaluate $\lim_{x \to -1} \frac{x^2 + 1}{(x^3 + 2)(x^4 + 1)}$
Step-by-Step Solution:
- Apply direct substitution of $x = -1$ into the expression: $\lim_{x \to -1} \frac{x^2 + 1}{(x^3 + 2)(x^4 + 1)} = \frac{(-1)^2 + 1}{((-1)^3 + 2)((-1)^4 + 1)}$
- Evaluate numerator: $(-1)^2 + 1 = 1 + 1 = 2$
- Evaluate denominator: $((-1)^3 + 2)((-1)^4 + 1) = (-1 + 2)(1 + 1) = (1)(2) = 2$
- Simplify final fraction: $\frac{2}{2} = 1$

A function $f(x)$ is continuous at a point $x = c$ if and only if all three of the following conditions are satisfied:
1. Limit Exists: $\lim_{x \to c} f(x)$ exists.
2. Function Value Exists: $f(c)$ exists (i.e., $f(c)$ is defined in the domain of $f$ ).
3. Limit Equals Function Value: $\lim_{x \to c} f(x) = f(c)$ .

Function: $f(x) = 3$ evaluated at $x = 2$
Condition 1 Verification: $\lim_{x \to 2} f(x) = \lim_{x \to 2} 3 = 3 \quad \checkmark$
Condition 2 Verification: $f(2) = 3 \quad \checkmark$
Condition 3 Verification: $\lim_{x \to 2} f(x) = f(2) \implies 3 = 3 \quad \checkmark$
Conclusion: $f(x) = 3$ is continuous at $x = 2$

Mathematical Definition:
- The limit exists at $x = c$ : $\lim_{x \to c} f(x)$ exists.
- The limit does not equal the function value at $x = c$ : $\lim_{x \to c} f(x) \neq f(c)$ (or $f(c)$ is undefined).
Graphical Property: The graph contains a hole at $x = c$ , with the function value either undefined or defined at a separate single point off the main curve.

Mathematical Definition:
- Both left-hand and right-hand limits exist at $x = c$ : $\lim_{x \to c^-} f(x)$ exists and $\lim_{x \to c^+} f(x)$ exists.
- The left-hand limit is not equal to the right-hand limit: $\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x)$ .
Graphical Property: The function breaks or "jumps" from one finite height to another at $x = c$

Left-Continuity: A function $f(x)$ is continuous from the left (left-continuous) at $x = c$ if: $\lim_{x \to c^-} f(x) = f(c)$
Right-Continuity: A function $f(x)$ is continuous from the right (right-continuous) at $x = c$ if: $\lim_{x \to c^+} f(x) = f(c)$

Function: $f(x) = \sqrt{x}$ evaluated at $x = 0$
Two-Sided Limit: $\lim_{x \to 0} \sqrt{x}$ Does Not Exist (DNE), because $\sqrt{x}$ is undefined for real numbers when $x < 0$
Right-Sided Limit: $\lim_{x \to 0^+} \sqrt{x} = 0 = f(0)$
Conclusion: $f(x) = \sqrt{x}$ is right-continuous at $x = 0$

A function is continuous on an interval if it is continuous at every interior point and possesses the appropriate one-sided continuity at any included endpoint.
Interval Classifications:
- Interval $[-4, -1)$ : Closed at $x = -4$ (right-continuous at $x = -4$ ), open at $x = -1$
- Interval $[-1, 1]$ : Closed at both endpoints $x = -1$ and $x = 1$ (right-continuous at $x = -1$ , left-continuous at $x = 1$ )
- Interval $(1, 3]$ : Open at $x = 1$ , closed at $x = 3$ (left-continuous at $x = 3$ )

Algebraic Combination Rules: If $f(x)$ and $g(x)$ are continuous at $x = c$ , then:
- Sum and Difference: $f(x) \pm g(x)$ is continuous at $x = c$
- Product: $f(x) \cdot g(x)$ is continuous at $x = c$
- Constant Multiple: $k \cdot f(x)$ is continuous at $x = c$ for any real constant $k$
- Quotient: $\frac{f(x)}{g(x)}$ is continuous at $x = c$ , provided $g(c) \neq 0$
- Composite Function: $(f \circ g)(x) = f(g(x))$ is continuous at $x = c$ if $g(x)$ is continuous at $c$ and $f(x)$ is continuous at $g(c)$
Continuity of Standard Function Classes:
- Polynomial Functions: Continuous everywhere on $(-\infty, \infty)$
- Rational Functions: Functions $\frac{P(x)}{Q(x)}$ are continuous everywhere in their domain (at all points where $Q(x) \neq 0$ )
- Trigonometric Functions: Continuous at all points in their respective domains (e.g., $\sin(x)$ and $\cos(x)$ are continuous for all real numbers)

Example 1: $f(x) = \cos(2x)$
- Continuous for all real numbers: $(-\infty, \infty)$
Example 2: $f(x) = \frac{x + 1}{x^2 - 4}$
- Find domain restrictions by setting denominator to zero: $x^2 - 4 = 0 \implies x^2 = 4 \implies x = \pm 2$
- Continuous at all real numbers except $x = \pm 2$ : $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$

To check whether a piecewise function is continuous at a boundary point $x = 0$ :
- Calculate the left-hand limit: $\lim_{x \to 0^-} f(x)$
- Calculate the right-hand limit: $\lim_{x \to 0^+} f(x)$
- Verify if $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$

Problem Setup: Given a system of equations derived by equating left-hand and right-hand limits at boundaries:
1. $-a + b = 2$
2. $2a + b = -2$
Step-by-Step Algebraic Solution:
- Subtract equation (2) from equation (1): $(-a + b) - (2a + b) = 2 - (-2)$ $-a - 2a = 4$ $-3a = 4$ $a = -\frac{4}{3}$
- Substitute $a = -\frac{4}{3}$ into equation (1) $-a + b = 2$ : $-\left(-\frac{4}{3}\right) + b = 2$ $\frac{4}{3} + b = 2$ $b = 2 - \frac{4}{3}$ $b = \frac{6}{3} - \frac{4}{3} = \frac{2}{3}$
Final Parameter Values:
- $a = -\frac{4}{3}$
- $b = \frac{2}{3}$