MAT131: Limits, Continuity, and Rates of Change Notes

Course Overview and Fundamental Topics

• Course Code: MAT131

• Lecture Date: September 2, 2026

• Core Curriculum Topics Covered:

  • Limits and Continuity: Mathematical definitions, distinction between function evaluation and limit evaluation, and graphical behavior.

  • Rates of Change: Average rate of change, secant lines, difference quotient formulation, and physical applications (velocity).

  • The Limit of a Function and One-Sided Limits: Formal definition of two-sided limits, right-hand limits, left-hand limits, and criteria for existence.

  • Calculating Limits Using the Limit Laws: Algebraic laws for limits, indeterminate forms, algebraic simplification, special trigonometric limits, and difference quotient calculations.

Average Rate of Change and Difference Quotients

• Definition of Difference Quotient:

  • The average rate of change of a function $f(x)$ over an interval from $x = a$ to $x = a + h$ is defined by the difference quotient:     $\text{Average Rate of Change} = \frac{f(a+h) - f(a)}{h}$

• Physical Application: Average Velocity:

  • Average velocity measures the total displacement of an object divided by the elapsed time interval $h$:     $\text{average velocity} = \frac{\text{displacement}}{\text{time}} = \frac{f(a+h) - f(a)}{h}$

• Geometric Interpretation:

  • Geometrically, the difference quotient represents the slope of the secant line connecting two points $P(a, f(a))$ and $Q(a+h, f(a+h))$ on the curve $y = f(x)$.

  • Horizontal Change (Run): $h = (a + h) - a$

  • Vertical Change (Rise): $f(a+h) - f(a)$

  • Secant Line Slope: $m_{\text{sec}} = \frac{f(a+h) - f(a)}{h}$

• Detailed Example 1: Motion of a Particle:

  • Problem: The displacement of a particle moving in a straight line is given by the equation of motion $f(t) = t^3 - 4t + 3$. Find the average velocity of the particle over the interval $0 \le t \le 4$.

  • Step-by-Step Solution:

  • Identify the starting time $a = 0$ and ending time $a + h = 4$, which gives time interval length $h = 4 - 0 = 4$.

  • Calculate displacement at initial time $t = 0$:       $f(0) = (0)^3 - 4(0) + 3 = 3$

  • Calculate displacement at final time $t = 4$:       $f(4) = (4)^3 - 4(4) + 3 = 64 - 16 + 3 = 51$

  • Compute total displacement:       $\Delta f = f(4) - f(0) = 51 - 3 = 48$

  • Calculate average velocity:       $\text{average velocity} = \frac{f(4) - f(0)}{4 - 0} = \frac{48}{4} = 12$

Definition of Limits and Basic Limit Properties

• Formal Definition of Limit:

  • The statement:     $\lim_{x \to c} f(x) = L$     means that $f(x)$ approaches the limit $L$ as $x$ approaches $c$.

  • Verbal reading: "The limit of $f(x)$, as $x$ approaches $c$, equals $L$."

• Basic Limit Properties:

  1. Constant Function Limit:

  • If $f$ is the constant function $f(x) = k$, then for any real value $c$:        $\lim_{x \to c} f(x) = \lim_{x \to c} k = k$

  1. Power/Polynomial Base Function Limit:

  • If $f$ is the polynomial power function $f(x) = x^n$, then for any real value $c$:        $\lim_{x \to c} f(x) = \lim_{x \to c} x^n = c^n$

Function Values vs. Existence of Limits

• Key Distinction:

  • There is a fundamental conceptual difference between:

  1. The function being defined at $x = a$ (i.e., whether $f(a)$ exists).

  2. The limit of $f(x)$ existing at $x = a$ (i.e., whether $\lim_{x \to a} f(x)$ exists).

  • Condition (1) relates directly to whether $f(x)$ is continuous at $x = a$

  • Condition (2) relates directly to the behavior of one-sided limits as $x$ approaches $a$ from left and right.

• Illustrative Rational Function Example:

  • Consider the function:     $f(x) = \frac{x^3 + 1}{x + 1}$

  • At $x = -1$, direct evaluation yields $f(-1) = \frac{(-1)^3 + 1}{-1 + 1} = \frac{0}{0}$, which is undefined. Hence, condition (1) fails.

  • However, factoring the sum of cubes in the numerator gives:     $x^3 + 1 = (x + 1)(x^2 - x + 1)$

  • For all $x \neq -1$:     $f(x) = \frac{(x + 1)(x^2 - x + 1)}{x + 1} = x^2 - x + 1$

  • Evaluating the limit as $x \to -1$:     $\lim_{x \to -1} f(x) = \lim_{x \to -1} (x^2 - x + 1) = (-1)^2 - (-1) + 1 = 1 + 1 + 1 = 3$

  • Thus, the limit exists and equals $3$, despite $f(-1)$ being undefined. Graphically, this corresponds to a removable discontinuity (hole) at point $(-1, 3)$.

One-Sided Limits and Limit Existence Criteria

• Right-Hand Limit:

  • The right-hand limit means that $x$ approaches $c$ from values strictly greater than $c$ ($x \to c^+$).

  • Denoted as:     $\lim_{x \to c^+} f(x) = L$

• Left-Hand Limit:

  • The left-hand limit means that $x$ approaches $c$ from values strictly less than $c$ ($x \to c^-$).

  • Denoted as:     $\lim_{x \to c^-} f(x) = L$

• Theorem: Existence of a Two-Sided Limit:

  • The two-sided limit of $f(x)$ as $x$ approaches $c$ is $L$ if and only if both one-sided limits exist and are equal to $L$:     $\lim_{x \to c} f(x) = L \iff \lim_{x \to c^+} f(x) = L \quad \text{and} \quad \lim_{x \to c^-} f(x) = L$

Functions and Limits Failing to Exist

• Common Mechanisms for Non-Existence of Limits:

  1. Jump Discontinuity (Differing One-Sided Limits):

  • Example function: $f(x) = \frac{|x|}{x}$

  • For $x > 0$, $|x| = x \implies f(x) = \frac{x}{x} = 1$. Thus, $\lim_{x \to 0^+} f(x) = 1$.

  • For $x < 0$, $|x| = -x \implies f(x) = \frac{-x}{x} = -1$. Thus, $\lim_{x \to 0^-} f(x) = -1$.

  • Since $\lim_{x \to 0^+} f(x) \neq \lim_{x \to 0^-} f(x)$, the limit $\lim_{x \to 0} \frac{|x|}{x}$ does not exist.

  1. Infinite Discontinuity (Unbounded Growth):

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

  • As $x \to 0^+$, $f(x) \to \infty$

  • As $x \to 0^-$, $f(x) \to -\infty$

  • Because values grow without bound and approach opposite infinities, no finite real limit exists.

  1. Oscillatory Behavior:

  • Example function: $f(x) = \sin\left(\frac{1}{x}\right)$

  • As $x \to 0$, the frequency of oscillation approaches infinity, forcing $f(x)$ to oscillate infinitely between $-1$ and $1$ in every open interval around $0$. It never approaches a single fixed value, so the limit fails to exist.

Graphical Analysis of Limits and Function Values

• Comprehensive Graphical Example:

• Evaluate limits and function values based on the graph of $y = f(x)$:

  • (a) $\lim_{x \to 3^-} f(x)$: Approaching $x = 3$ along the semicircular path from the left, the $y$-value approaches $0$.       $\lim_{x \to 3^-} f(x) = 0$

  • (b) $\lim_{x \to 3^+} f(x)$: Approaching $x = 3$ along the horizontal line segment from the right, the $y$-value approaches $3$.       $\lim_{x \to 3^+} f(x) = 3$

  • (c) $\lim_{x \to 3} f(x)$: Since $\lim_{x \to 3^-} f(x) = 0 \neq 3 = \lim_{x \to 3^+} f(x)$, the two-sided limit does not exist.       $\lim_{x \to 3} f(x) = \text{Does Not Exist (DNE)}$

  • (d) $\lim_{x \to 6} f(x)$: Approaching $x = 6$ from the left along the horizontal segment gives $y \to 3$. Approaching $x = 6$ from the right along the downward linear slope gives $y \to 3$. Since both one-sided limits equal $3$:       $\lim_{x \to 6} f(x) = 3$

  • (e) $f(3)$: Solid dot is located on the $x$-axis at $(3, 0)$.       $f(3) = 0$

  • (f) $f(6)$: Solid isolated point is defined at $(6, 1)$.       $f(6) = 1$

Limit Laws and Algebraic Manipulation

• Statement of Limit Laws:

  • Let $c$ and $k$ be real numbers, and assume that both limits $\lim_{x \to c} f(x)$ and $\lim_{x \to c} g(x)$ exist.

  1. Sum and Difference Law:      $\lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

  2. Product Law:      $\lim_{x \to c} [f(x) \cdot g(x)] = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x)$

  3. Constant Multiple Law:      $\lim_{x \to c} [k \cdot f(x)] = k \lim_{x \to c} f(x)$

  4. Quotient Law:      $\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)} \quad \text{provided} \quad \lim_{x \to c} g(x) \neq 0$

  5. Power Law:      $\lim_{x \to c} [f(x)]^n = \left[\lim_{x \to c} f(x)\right]^n$

  6. Root Law:      $\lim_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to c} f(x)} \quad (\text{assuming } \lim_{x \to c} f(x) > 0 \text{ if } n \text{ is even})$

• Algebraic Limit Examples with Piecewise & Absolute Value Functions:

  • Piecewise Function Example:

  • Given:       $f(x) = \begin{cases} x^2 + 3, & x \neq 1 \\ 1, & x = 1 \end{cases}$

  • Find $\lim_{x \to 1} f(x)$.

  • Solution: The limit as $x \to 1$ concerns values of $x$ near $1$, but not equal to $1$. Therefore, we evaluate using the branch $x^2 + 3$:       $\lim_{x \to 1} f(x) = \lim_{x \to 1} (x^2 + 3) = 1^2 + 3 = 4$

  • Note that $f(1) = 1 \neq 4$.

  • Absolute Value Rational Example:

  • Find $\lim_{x \to 1} \frac{|x - 1|}{1 - x}$.

  • Solution: Analyze one-sided limits:

    • For $x < 1$, $x - 1 < 0 \implies |x - 1| = -(x - 1) = 1 - x$:         $\lim_{x \to 1^-} \frac{1 - x}{1 - x} = \lim_{x \to 1^-} 1 = 1$

    • For $x > 1$, $x - 1 > 0 \implies |x - 1| = x - 1 = -(1 - x)$:         $\lim_{x \to 1^+} \frac{-(1 - x)}{1 - x} = \lim_{x \to 1^+} (-1) = -1$

    • Since $\lim_{x \to 1^-} \neq \lim_{x \to 1^+}$, $\lim_{x \to 1} \frac{|x - 1|}{1 - x}$ does not exist (DNE).

Special Trigonometric Limits and Applications

• Fundamental Trigonometric Limit Identities:

  1. $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$

  2. $\lim_{x \to 0} \frac{1 - \cos(x)}{x} = 0$

• Factoring and Conjugate Rationalization Worked Examples:

  • Example (a): Find $\lim_{x \to 2} \frac{x^2 + 3x - 10}{x - 2}$

  • Direct substitution gives $\frac{0}{0}$.

  • Factor numerator: $x^2 + 3x - 10 = (x + 5)(x - 2)$.

  • Simplify:       $\lim_{x \to 2} \frac{(x + 5)(x - 2)}{x - 2} = \lim_{x \to 2} (x + 5) = 2 + 5 = 7$

  • Example (b): Find $\lim_{x \to 1} \frac{\sqrt{x + 3} - 2}{x - 1}$

  • Direct substitution gives $\frac{0}{0}$.

  • Rationalize numerator by multiplying numerator and denominator by conjugate $\sqrt{x + 3} + 2$:       $\frac{(\sqrt{x + 3} - 2)(\sqrt{x + 3} + 2)}{(x - 1)(\sqrt{x + 3} + 2)} = \frac{(x + 3) - 4}{(x - 1)(\sqrt{x + 3} + 2)} = \frac{x - 1}{(x - 1)(\sqrt{x + 3} + 2)}$

  • Cancel $(x - 1)$ factor:       $\lim_{x \to 1} \frac{1}{\sqrt{x + 3} + 2} = \frac{1}{\sqrt{1 + 3} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$

• Trigonometric Limit Worked Examples:

  • Example (a): Find $\lim_{x \to 0} \frac{\sin(4x)}{3x}$

  • Rewrite expression to match $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$:       $\lim_{x \to 0} \left(\frac{4}{3} \cdot \frac{\sin(4x)}{4x}\right) = \frac{4}{3} \cdot 1 = \frac{4}{3}$

  • Example (b): Find $\lim_{x \to 0} \frac{\tan(x)}{x}$

  • Express $\tan(x) = \frac{\sin(x)}{\cos(x)}$:       $\lim_{x \to 0} \left(\frac{\sin(x)}{x} \cdot \frac{1}{\cos(x)}\right) = \left(\lim_{x \to 0} \frac{\sin(x)}{x}\right) \cdot \left(\lim_{x \to 0} \frac{1}{\cos(x)}\right) = 1 \cdot \frac{1}{1} = 1$

  • Example (c): Find $\lim_{x \to \pi/4} \frac{\tan(x) - 1}{\sin(x) - \cos(x)}$

  • Rewrite numerator: $\tan(x) - 1 = \frac{\sin(x)}{\cos(x)} - 1 = \frac{\sin(x) - \cos(x)}{\cos(x)}$

  • Substitute into fraction:       $\frac{\frac{\sin(x) - \cos(x)}{\cos(x)}}{\sin(x) - \cos(x)} = \frac{1}{\cos(x)}$

  • Evaluate limit as $x \to \frac{\pi}{4}$:       $\lim_{x \to \pi/4} \frac{1}{\cos(x)} = \frac{1}{\cos(\pi/4)} = \frac{1}{1/\sqrt{2}} = \sqrt{2}$

  • Example (d): Find $\lim_{x \to \pi/3} \frac{\sin\left(\frac{\pi}{3} - x\right)}{\frac{\pi}{3} - x}$

  • Perform variable substitution $u = \frac{\pi}{3} - x$. As $x \to \frac{\pi}{3}$, $u \to 0$.       $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$

  • Example (e): Find $\lim_{x \to 0} \frac{\sin(3x)}{\sin(2x)}$

  • Multiply and divide by $3x$ and $2x$:       $\frac{\sin(3x)}{\sin(2x)} = \frac{\frac{\sin(3x)}{3x} \cdot 3x}{\frac{\sin(2x)}{2x} \cdot 2x} = \frac{3}{2} \cdot \frac{\frac{\sin(3x)}{3x}}{\frac{\sin(2x)}{2x}}$

  • Take limit as $x \to 0$:       $\frac{3}{2} \cdot \frac{1}{1} = \frac{3}{2}$

  • Example (f): Find $\lim_{\theta \to 0} \frac{\theta + \theta \cos(\theta)}{\sin(\theta) \cos(\theta)}$

  • Factor $\theta$ in numerator:       $\frac{\theta (1 + \cos(\theta))}{\sin(\theta) \cos(\theta)} = \left(\frac{\theta}{\sin(\theta)}\right) \cdot \left(\frac{1 + \cos(\theta)}{\cos(\theta)}\right)$

  • Take limit as $\theta \to 0$:       $\left(\lim_{\theta \to 0} \frac{\theta}{\sin(\theta)}\right) \cdot \left(\lim_{\theta \to 0} \frac{1 + \cos(\theta)}{\cos(\theta)}\right) = 1 \cdot \frac{1 + 1}{1} = 2$

Advanced Limit Problem Calculations and Difference Quotient Limits

• Solving for Parameter $a$:

  • Problem: If $\lim_{x \to 0} \frac{\sqrt{2 + ax} - \sqrt{2}}{x} = \sqrt{2}$, find the value of parameter $a$

  • Solution Steps:

  • Rationalize numerator by multiplying numerator and denominator by conjugate $\sqrt{2 + ax} + \sqrt{2}$:       $\frac{(\sqrt{2 + ax} - \sqrt{2})(\sqrt{2 + ax} + \sqrt{2})}{x(\sqrt{2 + ax} + \sqrt{2})} = \frac{(2 + ax) - 2}{x(\sqrt{2 + ax} + \sqrt{2})} = \frac{ax}{x(\sqrt{2 + ax} + \sqrt{2})} = \frac{a}{\sqrt{2 + ax} + \sqrt{2}}$

  • Take limit as $x \to 0$:       $\lim_{x \to 0} \frac{a}{\sqrt{2 + ax} + \sqrt{2}} = \frac{a}{\sqrt{2} + \sqrt{2}} = \frac{a}{2\sqrt{2}}$

  • Set limit equal to given value $\sqrt{2}$:       $\frac{a}{2\sqrt{2}} = \sqrt{2} \implies a = 2\sqrt{2} \cdot \sqrt{2} = 2 \cdot 2 = 4$

• Difference Quotient Limit for Radical Function:

  • Problem: Find $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$, if $f(x) = \sqrt{2x + 1}$

  • Solution Steps:

  • Form difference quotient expression:       $\frac{f(x+h) - f(x)}{h} = \frac{\sqrt{2(x+h) + 1} - \sqrt{2x + 1}}{h} = \frac{\sqrt{2x + 2h + 1} - \sqrt{2x + 1}}{h}$

  • Multiply numerator and denominator by conjugate $\sqrt{2x + 2h + 1} + \sqrt{2x + 1}$:       $\frac{(\sqrt{2x + 2h + 1} - \sqrt{2x + 1})(\sqrt{2x + 2h + 1} + \sqrt{2x + 1})}{h(\sqrt{2x + 2h + 1} + \sqrt{2x + 1})} = \frac{(2x + 2h + 1) - (2x + 1)}{h(\sqrt{2x + 2h + 1} + \sqrt{2x + 1})}$

  • Simplify numerator:       $\frac{2h}{h(\sqrt{2x + 2h + 1} + \sqrt{2x + 1})} = \frac{2}{\sqrt{2x + 2h + 1} + \sqrt{2x + 1}}$

  • Evaluate limit as $h \to 0$:       $\lim_{h \to 0} \frac{2}{\sqrt{2x + 2h + 1} + \sqrt{2x + 1}} = \frac{2}{\sqrt{2x + 1} + \sqrt{2x + 1}} = \frac{2}{2\sqrt{2x + 1}} = \frac{1}{\sqrt{2x + 1}}$