VCE Mathematical Methods Units 1 & 2: Limits, Continuity, and Differentiability

Warm-Up Cyclical Revision: Secant Line Gradient Evaluation

• Problem Statement:   Find the gradient of the secant line between the points where $x = 1$ and $x = 4$ on the function $f(x) = \sqrt{x + 3}$.

  • Option A: $0.12$

  • Option B: $0.20$

  • Option C: $0.25$

  • Option D: $0.33$

• Step-by-Step Solution:

  • The gradient of a secant line between two points $x_1$ and $x_2$ is given by the average rate of change formula:     $\text{Gradient} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

  • Substituting $x_1 = 1$ and $x_2 = 4$:     $\text{Gradient} = \frac{f(4) - f(1)}{4 - 1}$

  • Calculate the individual function values:     $f(4) = \sqrt{4 + 3} = \sqrt{7}$     $f(1) = \sqrt{1 + 3} = \sqrt{4} = 2$

  • Substitute the function values back into the gradient formula:     $\text{Gradient} = \frac{\sqrt{7} - \sqrt{4}}{4 - 1} = \frac{\sqrt{7} - 2}{3}$

  • Evaluate numerically:     $\sqrt{7} \approx 2.645751$     $\text{Gradient} \approx \frac{2.645751 - 2}{3} = \frac{0.645751}{3} \approx 0.21525$

  • Conclusion: The calculated gradient of approximately $0.215$ is closest to Option B ($0.20$).

Algebraic Simplification Practice

• Task: Simplify the rational function expression $f(x) = \frac{3x^2 - 5x + 2}{x - 1}$.

• Step-by-Step Factorisation & Simplification:

  • Factorise the quadratic numerator $3x^2 - 5x + 2$:

  • Find two factors of $3 \times 2 = 6$ that add up to $-5$, which are $-3$ and $-2$.

  • Rewrite the quadratic term: $3x^2 - 3x - 2x + 2 = 3x(x - 1) - 2(x - 1) = (3x - 2)(x - 1)$.

  • Substitute the factored form back into the numerator:     $f(x) = \frac{(3x - 2)(x - 1)}{x - 1}$

  • Cancel out the common linear factor $(x - 1)$ for all $x \neq 1$:     $f(x) = 3x - 2, \quad \text{for } x \neq 1$

Fundamental Concepts and Key Vocabulary

• Limit: An operation determining the value that a function $f(x)$ approaches as the input $x$ approaches a specified target value $a$.

• Piecewise (Hybrid) Function: A function defined by multiple distinct sub-rules over different disjoint intervals of its domain.

• Discontinuous Function: A function whose graph contains breaks, gaps, holes, or vertical asymptotes. There are three primary types of discontinuities:

  1. Jump Discontinuity: Occurs when the left-hand limit and right-hand limit both exist as finite values but are not equal to each other ($\lim_{x \rightarrow a^-} f(x) \neq \lim_{x \rightarrow a^+} f(x)$).      

     

  2. Point Discontinuity (Removable Discontinuity): Occurs when the limit as $x \rightarrow a$ exists, but either $f(a)$ is undefined or $f(a) \neq \lim_{x \rightarrow a} f(x)$.      

     

  3. Infinite Discontinuity: Occurs when one or both of the one-sided limits tend toward positive or negative infinity ($\pm \infty$), producing a vertical asymptote at $x = a$.      

     

Mathematical Definition and Properties of Limits

• Core Definition:   Limits determine whether a limit exists at a specific point by analyzing the trend of function values near that point.

• Crucial Distinction:   Limits are concerned exclusively with the behaviour of a function as it approaches a particular point $x = a$, and NOT with its behaviour or defined value at the point $x = a$ itself.

• Formal Limit Notation:   $\lim_{x \rightarrow a} f(x) = L$

  • Statement Translation: "The limit of $f(x)$, as $x$ approaches $a$, is $L$".

  • Interpretation: "As $x$ approaches $a$, $f(x)$ approaches $L$".

• Direct Substitution Rule for Well-Behaved Functions:   In most standard cases where a function is continuous and well-behaved at $x = a$, the limit simply equals the evaluated function value:   $\lim_{x \rightarrow a} f(x) = f(a)$

Systematic Evaluation Techniques for Limits

• Procedure for Evaluating $\lim_{x \rightarrow a} f(x)$:

  1. Direct Substitution: Check whether the function is well-behaved at $x = a$. If direct substitution produces a defined real number, that number is the limit.

  2. Algebraic Simplification / Factorisation: If direct substitution at $x = a$ produces an indeterminate form such as $\frac{0}{0}$, factorise and simplify the algebraic expression to eliminate common factors causing zero in the denominator, creating a well-behaved function, then substitute $x = a$.

  3. Non-Existence Criteria: If it is impossible to simplify the expression into a well-behaved form, or if the left-hand limit does not equal the right-hand limit, the limit does not exist.

• Rule: Factorise before evaluation.

Algebra of Limits

Let $f(x)$ and $g(x)$ be functions whose limits as $x \rightarrow a$ exist, and let $c$ be any real number constant:

• Sum Rule:   $\lim_{x \rightarrow a} (f(x) + g(x)) = \lim_{x \rightarrow a} f(x) + \lim_{x \rightarrow a} g(x)$   The limit of a sum is equal to the sum of the individual limits.

• Constant Multiple Rule:   $\lim_{x \rightarrow a} (c \cdot f(x)) = c \cdot \lim_{x \rightarrow a} f(x)$   A constant scalar factor can be factored outside the limit operator.

• Product Rule:   $\lim_{x \rightarrow a} (f(x) \cdot g(x)) = \lim_{x \rightarrow a} f(x) \cdot \lim_{x \rightarrow a} g(x)$   The limit of a product is equal to the product of the individual limits.

• Quotient Rule:   $\lim_{x \rightarrow a} \left(\frac{f(x)}{g(x)}\right) = \frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)}, \quad \text{provided } \lim_{x \rightarrow a} g(x) \neq 0$   The limit of a quotient is equal to the quotient of the limits, provided the limit of the denominator is non-zero.

Practice Limit Evaluations (Worked Examples)

• Problem A: Calculate $\lim_{x \rightarrow 2} 4(x + 2)$

  • Applying limit laws:     $\lim_{x \rightarrow 2} 4(x + 2) = \lim_{x \rightarrow 2} (4) \times \lim_{x \rightarrow 2} (x + 2)$     $\lim_{x \rightarrow 2} 4(x + 2) = 4 \times (2 + 2) = 4 \times 4 = 16$

  • (Note: evaluating as $\lim_{x \rightarrow 2} (4x) \times \lim_{x \rightarrow 2} (x + 2) = 4(2) \times (2 + 2) = 8 \times 4 = 32$).

• Problem B: Calculate $\lim_{x \rightarrow 3} \left(\frac{5x + 2}{x - 2}\right)$

  • Applying the quotient rule directly:     $\lim_{x \rightarrow 3} \left(\frac{5x + 2}{x - 2}\right) = \frac{\lim_{x \rightarrow 3} (5x + 2)}{\lim_{x \rightarrow 3} (x - 2)} = \frac{5(3) + 2}{3 - 2} = \frac{15 + 2}{1} = 17$

• Problem C: Calculate $\lim_{x \rightarrow 5} \left(\frac{x^2 - 7x + 10}{x^2 - 25}\right)$

  • Direct substitution at $x = 5$ yields $\frac{5^2 - 7(5) + 10}{5^2 - 25} = \frac{0}{0}$ (indeterminate form).

  • Factorise numerator: $x^2 - 7x + 10 = (x - 2)(x - 5)$.

  • Factorise denominator: $x^2 - 25 = (x + 5)(x - 5)$.

  • Simplify by cancelling the common factor $(x - 5)$:     $\lim_{x \rightarrow 5} \left(\frac{(x - 2)(x - 5)}{(x + 5)(x - 5)}\right) = \lim_{x \rightarrow 5} \left(\frac{x - 2}{x + 5}\right)$

  • Substitute $x = 5$ into the simplified expression:     $\frac{5 - 2}{5 + 5} = \frac{3}{10}$

• Problem D: Calculate $\lim_{x \rightarrow 1} \left(\frac{1}{x - 1}\right)$

  • Direct substitution yields $\frac{1}{0}$, which is undefined.

  • The expression cannot be factorised or simplified into a well-behaved function.

  • As $x \rightarrow 1^+$, $\frac{1}{x-1} \rightarrow +\infty$; as $x \rightarrow 1^-$, $\frac{1}{x-1} \rightarrow -\infty$.

  • Conclusion: The function is neither well-behaved nor simplifiable. Therefore, $\lim_{x \rightarrow 1} \left(\frac{1}{x - 1}\right)$ does not exist.

One-Sided Limits (Left and Right Limits)

• Definitions:

  • Left-Hand Limit ($L^-$): The value that $f(x)$ approaches as $x$ approaches $a$ strictly from values less than $a$ (from the left):     $L^- = \lim_{x \rightarrow a^-} f(x)$

  • Right-Hand Limit ($L^+$): The value that $f(x)$ approaches as $x$ approaches $a$ strictly from values greater than $a$ (from the right):     $L^+ = \lim_{x \rightarrow a^+} f(x)$

• General Limit Existence Condition:   The overall limit $\lim_{x \rightarrow a} f(x)$ exists if and only if both one-sided limits exist and are equal:   $\lim_{x \rightarrow a} f(x) = L \iff \lim_{x \rightarrow a^-} f(x) = \lim_{x \rightarrow a^+} f(x) = L$   If $L^- \neq L^+$, the limit does not exist.

• Worked Graphical Example:   

  

  • Evaluate the limit of $f(x)$ as $x \rightarrow 2$:

  • Left-hand limit: $L^- = \lim_{x \rightarrow 2^-} f(x) = 4$

  • Right-hand limit: $L^+ = \lim_{x \rightarrow 2^+} f(x) = 6$

  • Since $L^- \neq L^+$ ($4 \neq 6$), the overall limit $\lim_{x \rightarrow 2} f(x)$ does not exist.

Criteria and Assessment of Function Continuity

• Definition of Continuity:   A function $f(x)$ is continuous at a point $x = a$ if its graph can be drawn through $x = a$ without lifting the pencil—meaning there is no break, jump, hole, or asymptote at that point.

• Three Mandatory Conditions for Continuity at $x = a$:

  1. $f(a)$ exists ($a$ is within the domain of $f$).

  2. $\lim_{x \rightarrow a} f(x)$ exists ($\lim_{x \rightarrow a^-} f(x) = \lim_{x \rightarrow a^+} f(x)$).

  3. $\lim_{x \rightarrow a} f(x) = f(a)$ (the value approached by the limit equals the defined function value at $x = a$).

• Types of Discontinuities:

  • Vertical asymptotes (infinite discontinuity)

  • Unjoined branches (jump discontinuity)

  • Missing points / holes (removable point discontinuity)

Worked Examples on Function Continuity

• Problem Statement:   Consider the function $f(x) = \frac{x^2 + x}{x + 1}$.

• Part a: State its maximal domain.

  • Setting denominator equal to zero: $x + 1 = 0 \implies x = -1$.

  • Maximal domain: $\mathbb{R} \setminus \{-1\}$.

• Part b: Calculate $\lim_{x \rightarrow -1} f(x)$.

  • Factorise numerator: $x^2 + x = x(x + 1)$.

  • Simplify expression:     $\lim_{x \rightarrow -1} f(x) = \lim_{x \rightarrow -1} \frac{x(x + 1)}{x + 1} = \lim_{x \rightarrow -1} x = -1$

• Part c: Explain why the function is not continuous at $x = -1$. logic

  • The maximal domain of $f$ is $\mathbb{R} \setminus \{-1\}$, which means $f(-1)$ is undefined / does not exist.

  • Because condition 1 of continuity fails ($f(-1)$ does not exist), $f(x)$ is not continuous at x = -1$.\n\n- **Part d: Sketch the graph of y = f(x).**\n  \n - The graph is a straight line y = x$with an open circle (hole) at the coordinate point$(-1, -1).\n\n# Technology-Active Evaluation Using CAS (TI-Nspire CX II)\n\n- **Problem Scenario**:\n  A piecewise function is defined as:\n  f(x) = \begin{cases} x, & x < 1 \ 1, & x = 1 \ x^2, & x > 1 \end{cases}\n\n- **CAS Step-by-Step Procedure**:\n 1. **Define the Function**:\n - On a Calculator page, press `MENU` -> `1: Actions` -> `1: Define`.\n - Access the piecewise template via `|{|` button, then select the 3-case piecewise template.\n - Enter: `Define f(x) = { x, x < 1 ; 1, x = 1 ; x^2, x > 1 }`\n - Press `ENTER`. CAS returns `Done`.\n 2. **Calculate the Limit**:\n - Press `MENU` -> `4: Calculus` -> `4: Limit`.\n - Enter line as: \lim_{x \rightarrow 1} (f(x))\n - Press `ENTER`. CAS output: `1`.\n 3. **Evaluate Function at Point**:\n - Enter line as: `f(1)`\n - Press `ENTER`. CAS output: `1`.\n 4. **Determine Continuity**:\n - Since \lim_{x \rightarrow 1} f(x) = 1$and$f(1) = 1, the limit value equals the function value.\n - Therefore, f(x)$is **continuous** at$x = 1$.

  1. Confirm Graphically:

  • Open a Graphs page.

  • Enter entry line as: $f1(x) = f(x)$

  • Press ENTER. The graph shows a smooth continuous curve passing through $(1, 1)$.

Concepts and Criteria of Differentiability

• Geometric Concept:   Drawing a tangent line at a point $x = a$ allows determination of the gradient of the function at that exact point.   

  

• Formal Definition of Differentiability:   A function $f(x)$ is differentiable at $x = a$ if and only if:

  1. $f(x)$ is continuous at $x = a$.

  2. The derivative from the left equals the derivative from the right at $x = a$:      $f'(a^-) = f'(a^+)$

• Critical Notes & Caveats:

  • Continuity is a necessary condition for differentiability, but not a sufficient condition (i.e., not all continuous functions are differentiable).

  • Functions are not differentiable at sharp corners, cusps, vertical tangents, or points of discontinuity.

  • To test for differentiability algebraically at boundary points, verify whether $f'(a^-) = f'(a^+)$.

Worked Examples on Testing Differentiability

• Problem Statement:   The function defined as $f(x) = \begin{cases} x, & x \le 1 \\ x^2, & x > 1 \end{cases}$ is continuous at x = 1$.\n\n- **Part a: Test whether the function is differentiable at x = 1.**\n - **Derivative from the left (x \le 1)**:\n    f(x) = x \implies f'(x) = 1\n    Evaluating at x = 1:\n    f'(1^-) = 1\n - **Derivative from the right (x > 1)**:\n    f(x) = x^2 \implies f'(x) = 2x\n    Evaluating at x = 1:\n    f'(1^+) = 2(1) = 2\n - **Comparison**:\n    Since f'(1^-) eq f'(1^+)$($1 eq 2$), the rate of change is discontinuous at$x = 1$.

  • Conclusion: The function is not differentiable at x = 1$.\n\n- **Part b: Give the rule f'(x)$stating its domain and sketch the graph of$y = f'(x).**\n - **Derivative Rule**:\n    f'(x) = \begin{cases} 1, & x < 1 \ 2x, & x > 1 \end{cases}\n - **Domain of Derivative**:\n    \mathbb{R} \setminus {1}$(or$(-\infty, 1) \cup (1, \infty)$). Note that$x = 1 must be excluded.\n - **Graph of Derivative y = f'(x)**:\n    \n - For x < 1$: Horizontal line$y = 1$ending at an open circle at$(1, 1).\n - For x > 1$: Ray$y = 2x$starting at an open circle at$(1, 2)$$ and sloping upward.

Coursework Exercises and Assessment Requirements

• Textbook: Jacaranda Maths Quest Mathematical Methods 11 VCE Units 1 and 2 (Third Edition)

• Exercise Section: Exercise 12.2 – Limits, Continuity & Differentiability

• Technology-Free Questions: 1, 2, 3, 4, 6a, 6c, 6e, 7, 8, 9, 11, 12, 14

• Technology-Active Questions: 15, 16, 17, 18, 19

• Homework: Complete all uncompleted classwork and Exam Questions at the end of Exercise 12.2.

Self-Assessment and Mastery Criteria

• Level 1 (Just Starting): Able to evaluate the limit of a well-behaved function via direct substitution.

• Level 2 (Getting There): Able to evaluate the limit of a non-well-behaved function by factorising and simplifying prior to substitution.

• Level 3 (Almost There): Able to recognise when a limit does not exist for a function (e.g., when left and right limits are unequal or expressions are unbounded).

• Level 4 (Got It!): Able to evaluate the continuity of a function across all three formal conditions and rigorously test whether a continuous function is differentiable.