Calculus Continuity and Limits Vocabulary

Mathematical Definition and Conditions of Continuity

• Intuitive Concept:

  • A function is continuous over an interval if its graph can be drawn without lifting the writing instrument from the paper.

  • Lifting the pencil signifies a point of discontinuity, such as a jump, gap, hole, or asymptote.

• Formal Definition:

  • A function $f(x)$ is continuous at a point $x = a$ if and only if:         $\lim_{x \to a} f(x) = f(a)$

• Three Necessary Conditions for Continuity:

  • Condition 1: Function Value Exists:

    • $f(a)$ must be defined.

    • Visually represented by a solid filled dot at $x = a$.

    • There can be no division by zero, undefined values, or empty holes at $x = a$.

  • Condition 2: Two-Sided Limit Exists:

    • $\lim_{x \to a} f(x)$ must exist.

    • This requires the left-hand limit to equal the right-hand limit:             $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$

    • The function must approach the same $y$-value from both the left and right directions without experiencing conflicting directional behavior.

  • Condition 3: Limit Equals Function Value:

    • The limit as $x$ approaches $a$ must equal the function value evaluated at $a$:             $\lim_{x \to a} f(x) = f(a)$

• Failure of Conditions:

  • If any of the three conditions are not met at $x = a$, the function $f(x)$ is discontinuous at $x = a$.

Classifications of Discontinuities

• Removable Discontinuities:

  • Occur when $\lim_{x \to a} f(x)$ exists, but $f(a)$ is either undefined (a hole) or does not equal the limit.

  • Typically arise in rational functions containing a common factor in both the numerator and the denominator.

  • Example:         $f(x) = \frac{x^2 - 1}{x - 1}$         Factoring the numerator yields:         $f(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \quad \text{for } x \neq 1$         This graph represents a line $y = x + 1$ with a removable hole at $x = 1$.

• Nonremovable Discontinuities:

  • Jump Discontinuity:

    • Occur when the left-hand limit and right-hand limit both exist as finite values but are not equal:             $\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$

    • Visually represented by a physical gap between two separate branches of the graph.

  • Infinite Discontinuity (Asymptote):

    • Occurs when the function approaches positive or negative infinity as $x$ approaches $a$ from either side.

    • Visually represented by vertical asymptotes.

• Precedence / Hierarchy Rule between Discontinuities:

  • A jump discontinuity trumps a hole.

  • If a point $x = a$ displays both a gap between branches and an open hole or solid dot at the gap boundary, the discontinuity is classified as a jump discontinuity, not a removable discontinuity.

  • Example: If a graph contains removable holes at $x = 1$ and $x = 3$, but exhibits a branch gap at $x = 2$ containing an open hole and a dot, the discontinuity at $x = 2$ is strictly a jump discontinuity.

• Visualizing Discontinuities in Online Software (WebAssign):

  • Vertical asymptotes are frequently displayed as open space between graph branches without explicit dashed vertical lines.

  • Arrows on graph curves denote infinite extension toward asymptotes or boundaries.

One-Sided Continuity

• Continuous from the Right:

  • A function $f(x)$ is continuous from the right at $x = a$ if:         $\lim_{x \to a^+} f(x) = f(a)$

• Continuous from the Left:

  • A function $f(x)$ is continuous from the left at $x = a$ if:         $\lim_{x \to a^-} f(x) = f(a)$

• Determining One-Sided Continuity from Algebraic Piecewise Definitions:

  • In inequality notation, non-strict inequality symbols ($\le$ or $\ge$) indicate that the boundary point is included in the domain (a solid dot at $x = a$).

  • A solid dot guarantees one-sided continuity from the direction specified by the inequality.

  • Strict inequality symbols (< or >) indicate an open hole at the endpoint, meaning the function is not continuous at that endpoint from that direction.

  • Example Analysis:

    • For $x \le 1$, the presence of $\le$ means the graph has a solid dot at $x = 1$ when approaching from the left, establishing continuity from the left at $x = 1$.

    • For $2 \le x$, the presence of $\le$ means the graph has a solid dot at $x = 2$ when approaching from the right, establishing continuity from the right at $x = 2$.

Graph Transformations for Radical Functions

• Parent Square Root Function:

  • $y = \sqrt{x}$ starts at the origin $(0, 0)$ and extends into the first quadrant.

• Horizontal Shift:

  • $y = \sqrt{x - 2}$ represents a horizontal translation of the parent graph $y = \sqrt{x}$ shifted right by $2$ units, starting at $(2, 0)$.

  • Subtracting a constant inside the radical shifts the graph to the right; adding a constant inside shifts the graph to the left.

• Vertical Shift:

  • $y = \sqrt{x} - 2$ represents a vertical translation shifted down by $2$ units, starting at $(0, -2)$.

Continuity on Intervals and Notation

• Definition:

  • A function $f(x)$ is continuous on an interval if it is continuous at every real number inside that interval.

  • At the endpoints of a closed interval $[a, b]$, $f(x)$ must be continuous from the right at $a$ and continuous from the left at $b$.

• Interval Notation Rules:

  • Use square brackets [ or ] to indicate that an endpoint is included (solid dot, $\le$, or $\ge$).

  • Use round parentheses ( or ) to indicate that an endpoint is excluded (open hole, strict inequality, or vertical asymptote).

  • Positive infinity ($\infty$) and negative infinity ($-\infty$) always take round parentheses ( or ).

  • Sub-intervals are written separated by commas or union symbols ($\cup$).

• Constructing Intervals of Continuity from Piecewise Graphs:

  • Leftward infinite branch ending at $x = 1$ with a solid dot: Written as $(-\infty, 1]$.

  • Middle segment between $x = 1$ and $x = 2$ with open holes at both endpoints: Written as $(1, 2)$.

  • Rightward infinite branch starting at $x = 2$ with a solid dot: Written as $[2, \infty)$.

  • Branch with vertical asymptote at $x = 4$ and endpoint at $x = 6$: Written as $(4, 6)$.

Properties and Algebraic Combinations of Continuous Functions

• Algebraic Combinations:

  • If functions $f(x)$ and $g(x)$ are continuous at $x = a$, then the following functions are also continuous at $x = a$:

    1. Sum: $f(x) + g(x)$

    2. Difference: $f(x) - g(x)$

    3. Constant Multiple: $c \cdot f(x)$ for any real constant $c$

    4. Product: $f(x) \cdot g(x)$

    5. Quotient: $\frac{f(x)}{g(x)}$, provided $g(a) \neq 0$

• Composite Functions:

  • A composite function $(f \circ g)(x) = f(g(x))$ combines an inner function $g(x)$ and an outer function $f(x)$.

  • Example: If $f(x) = \sqrt{x}$ and $g(x) = x + 1$, the composite function is $f(g(x)) = \sqrt{x + 1}$.

Domains and Continuity of Standard Function Classes

• Fundamental Theorem of Continuous Functions:

  • The following function types are continuous at every number in their respective domains.

  • Finding the domain of these functions directly determines where they are continuous.

• Polynomial Functions:

  • Form: Expressions such as $f(x) = x^2 - 2x + 1$ or $f(x) = x^3 - 2$.

  • Domain: All real numbers, $(-\infty, \infty)$.

  • Continuity: Continuous everywhere on $(-\infty, \infty)$.

• Trigonometric Functions:

  • $\sin(x)$ and $\cos(x)$ have domains of $(-\infty, \infty)$ and are continuous everywhere.

  • $\tan(x)$ has domain restrictions at odd multiples of $\frac{\pi}{2}$ where vertical asymptotes occur.

• Rational Functions:

  • Form: $f(x) = \frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials.

  • Method: Set denominator equal to zero ($Q(x) = 0$) and solve for $x$ to identify values excluded from the domain.

  • Example 1:         $f(x) = \frac{1}{x^2 - 1}$         Set denominator to zero:         $x^2 - 1 = 0 \implies x^2 = 1 \implies x = \pm 1$         Excluded values are $x = -1$ and x = 1$.\n        Interval of continuity: (-\infty, -1) \cup (-1, 1) \cup (1, \infty).\n * *Example 2*:\n        f(x) = \frac{1}{3x - 5}\n        Set denominator to zero:\n        3x - 5 = 0 \implies x = \frac{5}{3}\n        Excluded value is x = \frac{5}{3}$.         Interval of continuity: $(-\infty, \frac{5}{3}) \cup (\frac{5}{3}, \infty)$.

• Root / Radical Functions:

  • Method: For even root functions (e.g., $\sqrt{g(x)}$), set the radicand greater than or equal to zero ($g(x) \ge 0$) and solve for x$.\n * *Example*:\n        f(x) = \sqrt{x - 2}\n        Set radicand:\n        x - 2 \ge 0 \implies x \ge 2\n        Interval of continuity: [2, \infty).\n\n* **Combination Radical and Rational Functions**:\n * *Method*: Analyze denominator restrictions and radical restrictions separately, then find their intersection on a number line.\n * *Example*:\n        f(x) = \frac{\sqrt{x}}{x^2 - 4}\n 1. Denominator restriction:\n           x^2 - 4 = 0 \implies x = \pm 2\n 2. Radicand restriction:\n           x \ge 0\n 3. Intersection:\n           x \ge 0$excluding$x = 2$(note that$x = -2$is already excluded by$x \ge 0).\n 4. Interval of continuity:\n           [0, 2) \cup (2, \infty)\n\n\n# Limit of Composite Functions Theorem\n\n* **Theorem Definition**:\n * If f$is continuous at$b$and$\lim_{x \to a} g(x) = b, then:\n        \lim_{x \to a} f(g(x)) = f\left(\lim_{x \to a} g(x)\right) = f(b)\n\n* **Step-by-Step Problem Walkthrough**:\n * *Given*: Function f$is continuous at$5$,$f(5) = 8$, and$f(4) = 3$(where$f(4) = 3 is extraneous distractor information).\n * *Task*: Evaluate \lim_{x \to 2} f(4x^2 - 11).\n * *Step 1*: Evaluate the limit of the inner function as x \to 2:\n        \lim_{x \to 2} (4x^2 - 11) = 4(2)^2 - 11 = 4(4) - 11 = 16 - 11 = 5\n * *Step 2*: Substitute the inner result 5$into the outer continuous function$f:\n        f(5) = 8\n * *Step 3*: Evaluate the limit of the resulting constant:\n        \lim_{x \to 2} 8 = 8\n\n\n# Evaluating Limits Using Continuity Properties\n\n* **Direct Substitution Method**:\n * If a function f(x)$is continuous at$x = a$, the limit as$x \to a can be evaluated by direct substitution:\n        \lim_{x \to a} f(x) = f(a)\n\n* **Step-by-Step Problem Walkthrough**:\n * *Task*: Evaluate \lim_{x \to 0} 4\cos(-1 + \cos(x)).\n * *Step 1*: Apply direct substitution for x = 0 inside the trigonometric function:\n        \cos(0) = 1\n * *Step 2*: Evaluate the inner argument:\n        -1 + \cos(0) = -1 + 1 = 0\n * *Step 3*: Evaluate the outer trigonometric expression:\n        4\cos(0) = 4(1) = 4\n\n\n# Intermediate Value Theorem (IVT)\n\n* **Theorem Statement**:\n * Let f(x)$be continuous on the closed interval$[a, b].\n * Let N$be any real number strictly between$f(a)$and$f(b)$, where$f(a) \neq f(b).\n * Then there exists at least one number c$in the open interval$(a, b) such that:\n        f(c) = N\n\n* **Conceptual Meaning**:\n * A continuous function cannot jump over any y$-values; it takes on every intermediate$y$-value between$f(a)$and$f(b).\n * Because the curve is smooth and continuous without gaps or asymptotes, every output value N$between$f(a)$and$f(b)$corresponds to at least one input value$c$in$(a, b).\n\n\n# Finding Parameters to Ensure Global Continuity of Piecewise Functions\n\n* **Problem Type 1: Determining Two Parameters (a$and$b)**\n * *Given*: Find values for constants a$and$b$such that$f(x) is continuous everywhere:\n        f(x) = \begin{cases} x^2 & \text{if } x < 2 \ ax^2 - bx + 3 & \text{if } 2 \le x < 3 \ 2x - a + b & \text{if } x \ge 3 \end{cases}\n\n * *Step 1: Enforce Continuity at the Boundary x = 2*:\n * Left-hand limit as x \to 2^-:\n            \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2) = (2)^2 = 4\n * Right-hand limit as x \to 2^+:\n            \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (ax^2 - bx + 3) = a(2)^2 - b(2) + 3 = 4a - 2b + 3\n * Equate left-hand limit and right-hand limit:\n            4a - 2b + 3 = 4 \implies 4a - 2b = 1 \quad \text{(Equation 1)}\n\n * *Step 2: Enforce Continuity at the Boundary x = 3*:\n * Left-hand limit as x \to 3^-:\n            \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (ax^2 - bx + 3) = a(3)^2 - b(3) + 3 = 9a - 3b + 3\n * Right-hand limit as x \to 3^+:\n            \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (2x - a + b) = 2(3) - a + b = 6 - a + b\n * Equate left-hand limit and right-hand limit:\n            9a - 3b + 3 = 6 - a + b\n            10a - 4b = 3 \quad \text{(Equation 2)}\n\n * *Step 3: Solve the System of Linear Equations*:\n * System:\n            \begin{cases} 4a - 2b = 1 \ 10a - 4b = 3 \end{cases}\n * Multiply Equation 1 by -2:\n            -2(4a - 2b = 1) \implies -8a + 4b = -2\n * Add the modified equation to Equation 2:\n            (-8a + 4b) + (10a - 4b) = -2 + 3\n            2a = 1 \implies a = \frac{1}{2}\n * Substitute a = \frac{1}{2} into Equation 1:\n            4\left(\frac{1}{2}\right) - 2b = 1 \implies 2 - 2b = 1 \implies -2b = -1 \implies b = \frac{1}{2}\n * *Result*: a = \frac{1}{2}$,$b = \frac{1}{2}.\n\n* **Problem Type 2: Determining One Parameter (c)**\n * *Given*: For what value of the constant c$is$f(x) continuous everywhere?\n        f(x) = \begin{cases} cx^2 + 2x & \text{if } x < 5 \ x^3 - cx & \text{if } x \ge 5 \end{cases}\n\n * *Step 1: Enforce Continuity at Boundary x = 5*:\n * Left-hand limit as x \to 5^-:\n            \lim_{x \to 5^-} (cx^2 + 2x) = c(5)^2 + 2(5) = 25c + 10\n * Right-hand limit as x \to 5^+:\n            \lim_{x \to 5^+} (x^3 - cx) = (5)^3 - c(5) = 125 - 5c\n\n * *Step 2: Equate Limits and Solve for c*:\n        25c + 10 = 125 - 5c\n        30c = 115\n        c = \frac{115}{30}\n\n * *Note on Software Input Formatting*:\n * In WebAssign homework submissions, fractions do not require simplification to lowest terms; entering \frac{115}{30}$is accepted alongside the simplified form$\frac{23}{6}$$.