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Limits and Derivatives

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Chapter Notes

Overview
This chapter introduces the fundamental concepts of calculus: limits and derivatives. It begins with an intuitive idea of a derivative as an instantaneous rate of change, using the example of velocity of a falling body. The concept of a limit is then formally introduced, including left-hand and right-hand limits, and the algebra of limits is established. The chapter culminates in the definition of the derivative of a function at a point and the derivatives of standard functions. The one big idea a student must walk away with is that the derivative measures the instantaneous rate of change of a function, which is the limit of the average rate of change as the interval shrinks to zero.
Key Concepts (10)
Limit of a function
As x approaches a, f(x) approaches l, then l is called the limit of the function f(x), symbolically written as lim x→a f(x) = l.
e.g. As x approaches 0, the value of f(x) = x² approaches 0, so lim x→0 x² = 0.
Left hand limit
The expected value of f at x = a given the values of f near x to the left of a, denoted as lim x→a⁻ f(x).
e.g. For f(x) = {1, x≤0; 2, x>0}, the left hand limit at 0 is 1.
Right hand limit
The expected value of f at x = a given the values of f near x to the right of a, denoted as lim x→a⁺ f(x).
e.g. For f(x) = {1, x≤0; 2, x>0}, the right hand limit at 0 is 2.
Existence of limit
If the right hand limit and left hand limit coincide, that common value is called the limit of f(x) at x = a.
e.g. For f(x) = x + 10, lim x→5⁻ f(x) = 15 and lim x→5⁺ f(x) = 15, so lim x→5 f(x) = 15.
Derivative of f at a
The derivative of f at a is defined by f'(a) = lim h→0 [f(a+h) - f(a)] / h, provided this limit exists.
e.g. For f(x) = 3x, f'(2) = lim h→0 [3(2+h) - 3(2)] / h = 3.
Instantaneous velocity
The velocity of a body at a particular instant of time, equal to the slope of the tangent to the distance-time curve at that point.
e.g. For s = 4.9t², the instantaneous velocity at t = 2 is between 19.551 m/s and 19.649 m/s.
Average velocity
Distance travelled between two time instants divided by the time interval.
e.g. Average velocity between t = 1 and t = 2 for s = 4.9t² is (19.6 - 4.9) / (2-1) = 14.7 m/s.
Polynomial function
A function f of degree n given by f(x) = a₀ + a₁x + a₂x² + ... + aₙxⁿ, where aᵢ are real numbers and aₙ ≠ 0.
e.g. f(x) = x³ - x² + 1 is a polynomial function of degree 3.
Rational function
A function f(x) = g(x)/h(x), where g(x) and h(x) are polynomials and h(x) ≠ 0.
e.g. f(x) = (x² - 4)/(x - 2) is a rational function.
Sandwich Theorem
Let f, g, h be real functions such that f(x) ≤ g(x) ≤ h(x) for all x in the common domain. If lim x→a f(x) = l = lim x→a h(x), then lim x→a g(x) = l.
e.g. Since cos x < sin x/x < 1 for 0 < x < π/2, and lim x→0 cos x = 1, by Sandwich Theorem lim x→0 sin x/x = 1.
Formulas & Equations (10)
lim x→a [f(x) + g(x)] = lim x→a f(x) + lim x→a g(x)
Reads as: Limit of sum of two functions is sum of the limits of the functions
Variables: f and g are functions; a is the point being approached
Use when: When finding the limit of a sum of two functions whose individual limits exist
⚠ Students forget to check that both individual limits exist before applying the rule.
lim x→a [f(x) - g(x)] = lim x→a f(x) - lim x→a g(x)
Reads as: Limit of difference of two functions is difference of the limits of the functions
Variables: f and g are functions; a is the point being approached
Use when: When finding the limit of a difference of two functions whose individual limits exist
⚠ Students incorrectly apply this when limits do not exist.
lim x→a [f(x) . g(x)] = lim x→a f(x) . lim x→a g(x)
Reads as: Limit of product of two functions is product of the limits of the functions
Variables: f and g are functions; a is the point being approached
Use when: When finding the limit of a product of two functions whose individual limits exist
⚠ Students forget to multiply the limits, instead trying to substitute directly.
lim x→a [f(x)/g(x)] = [lim x→a f(x)] / [lim x→a g(x)], provided lim x→a g(x) ≠ 0
Reads as: Limit of quotient of two functions is quotient of the limits of the functions, provided denominator limit is non-zero
Variables: f and g are functions; a is the point being approached
Use when: When finding the limit of a rational function where the denominator limit is non-zero
⚠ Students apply this rule even when lim x→a g(x) = 0, leading to undefined expressions.
lim x→a [λ f(x)] = λ lim x→a f(x)
Reads as: Limit of a constant times a function is the constant times the limit of the function
Variables: λ is a real number; f is a function; a is the point being approached
Use when: When a constant factor is multiplied with a function whose limit exists
⚠ Students forget to multiply the constant after finding the limit.
lim x→a xⁿ = aⁿ
Reads as: Limit of x to the power n as x approaches a is a to the power n
Variables: n is a natural number; a is the point being approached
Use when: When finding the limit of a simple power function
⚠ Students incorrectly apply this for negative or fractional n without checking conditions.
lim x→a (xⁿ - aⁿ)/(x - a) = n aⁿ⁻¹
Reads as: Limit of x to the n minus a to the n over x minus a as x approaches a is n times a to the n minus one
Variables: n is any positive integer; a is a real number
Use when: When evaluating limits of the form 0/0 involving difference of powers
⚠ Students forget the factor n and write just aⁿ⁻¹.
lim x→0 sin x / x = 1
Reads as: Limit of sine x over x as x approaches zero is one
Variables: x is in radians
Use when: When evaluating limits involving sine functions as the argument approaches zero
⚠ Students use degrees instead of radians, or apply the formula when x does not approach 0.
lim x→0 (1 - cos x) / x = 0
Reads as: Limit of one minus cosine x over x as x approaches zero is zero
Variables: x is in radians
Use when: When evaluating limits involving cosine functions as the argument approaches zero
⚠ Students incorrectly think this limit is 1 or use degrees.
f'(a) = lim h→0 [f(a+h) - f(a)] / h
Reads as: The derivative of f at a is the limit as h approaches zero of f of a plus h minus f of a over h
Variables: f is a real-valued function; a is a point in its domain; h is a small increment
Use when: When finding the derivative of a function at a specific point using the first principle
⚠ Students forget to write the limit sign or incorrectly substitute h=0 directly.
How It Works
Finding limit of a polynomial function
  1. Identify the polynomial function f(x) = a₀ + a₁x + a₂x² + ... + aₙxⁿ.
  2. Use the property lim x→a x = a and the algebra of limits to find lim x→a f(x).
  3. The limit is simply f(a), i.e., substitute x = a directly into the polynomial.
💡 For polynomial functions, the limit as x approaches a is always equal to the value of the function at a, provided the function is defined at a.
Finding limit of a rational function when denominator is non-zero at the point
  1. Identify the rational function f(x) = g(x)/h(x).
  2. Check if h(a) ≠ 0.
  3. If h(a) ≠ 0, then lim x→a f(x) = g(a)/h(a).
💡 This direct substitution only works when the denominator is not zero at the point.
Finding limit of a rational function when both numerator and denominator are zero at the point (0/0 form)
  1. Factorize both numerator g(x) and denominator h(x).
  2. Cancel the common factor (x - a) from numerator and denominator.
  3. Evaluate the limit of the simplified rational function by direct substitution.
💡 The cancellation is valid because x ≠ a in the limiting process.
Finding derivative using first principle
  1. Write the expression for f(a+h) by substituting (a+h) into the function.
  2. Compute f(a+h) - f(a).
  3. Divide the result by h.
  4. Take the limit as h → 0 of the resulting expression.
💡 The limit must exist for the derivative to exist. Do not substitute h=0 before taking the limit.
Evaluating limit of sin x / x as x → 0 using Sandwich Theorem
  1. Recall the inequality cos x < sin x/x < 1 for 0 < x < π/2.
  2. Note that lim x→0 cos x = 1 and lim x→0 1 = 1.
  3. By the Sandwich Theorem, lim x→0 sin x/x = 1.
💡 This proof only works for x in radians. The inequality is derived from geometric areas.
Exam Traps — Don't Make These Mistakes
⚠ Direct substitution in 0/0 form
✗ lim x→2 (x² - 4)/(x - 2) = (4 - 4)/(2 - 2) = 0/0 = 0
→ 0/0 is an indeterminate form, not a number. Direct substitution is invalid. The limit must be evaluated by factoring and cancelling the common factor.
✓ lim x→2 (x² - 4)/(x - 2) = lim x→2 (x-2)(x+2)/(x-2) = lim x→2 (x+2) = 4
⚠ Confusing left-hand and right-hand limits
✗ For f(x) = {1, x≤0; 2, x>0}, lim x→0 f(x) = 1
→ The left-hand limit is 1, but the right-hand limit is 2. Since they are different, the limit does not exist. The student only considered one side.
✓ lim x→0⁻ f(x) = 1, lim x→0⁺ f(x) = 2, so lim x→0 f(x) does not exist.
⚠ Using degrees instead of radians for trigonometric limits
✗ lim x→0 sin x / x = 1, where x is in degrees
→ The standard limit lim x→0 sin x / x = 1 is valid only when x is measured in radians. In degrees, the limit would be π/180.
✓ Ensure x is in radians. If the problem gives x in degrees, convert to radians first.
⚠ Forgetting the limit sign in derivative calculation
✗ f'(a) = [f(a+h) - f(a)] / h
→ The derivative is defined as the limit of the difference quotient as h→0, not the quotient itself. Omitting the limit sign is a conceptual error.
✓ f'(a) = lim h→0 [f(a+h) - f(a)] / h
⚠ Assuming limit equals function value always
✗ For f(x) = {x², x≠1; 0, x=1}, lim x→1 f(x) = f(1) = 0
→ The limit as x→1 is determined by values near 1, not at 1. Since f(x) = x² for x≠1, the limit is 1, not 0. The function value and limit can be different.
✓ lim x→1 f(x) = lim x→1 x² = 1, even though f(1) = 0.
Compare & Contrast
Direction of approach
Left-hand limit
Right-hand limit
Left-hand limit considers values of x approaching a from the left (x < a), while right-hand limit considers values approaching from the right (x > a).
Time interval
Average velocity
Instantaneous velocity
Average velocity is the change in position over a finite time interval, while instantaneous velocity is the limit of average velocity as the time interval approaches zero, representing the velocity at a single instant.
Dependence on nearby vs. exact point
Limit of a function at a point
Value of the function at that point
The limit is the value the function approaches as x gets arbitrarily close to a, while the function value is the actual output at x = a. They may be equal or different.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA (3 marks) on limits, 1 LA (5 marks) on derivatives or a combination of limits and derivatives
Most asked: Evaluation of limits using algebraic manipulation (factorization, rationalization) · Evaluation of limits using standard trigonometric limits (sin x/x, (1-cos x)/x) · Finding derivatives using the first principle (definition) · Sandwich Theorem based problems
Don't skip: Finding derivative of a function using first principle (5 marks); Evaluating limits of rational functions in 0/0 form (3-5 marks); Limits involving trigonometric functions and Sandwich Theorem (3-5 marks)
One-Line Revision (read aloud before the exam)
  1. The limit of a function at a point is the value it approaches as the input gets arbitrarily close to that point.
  2. A limit exists only if the left-hand limit and right-hand limit are equal.
  3. For polynomial functions, the limit as x→a is simply f(a).
  4. For rational functions, if denominator is zero at the point, factor and cancel common factors before substituting.
  5. The derivative of f at a is defined as f'(a) = lim h→0 [f(a+h) - f(a)] / h.
  6. The derivative represents the instantaneous rate of change of a function.
  7. Instantaneous velocity is the derivative of the distance function with respect to time.
  8. The standard limit lim x→0 sin x / x = 1 is valid only when x is in radians.
  9. The standard limit lim x→0 (1 - cos x) / x = 0 is derived using the identity 1 - cos x = 2 sin²(x/2).
  10. The Sandwich Theorem states that if f(x) ≤ g(x) ≤ h(x) and lim f(x) = lim h(x) = l, then lim g(x) = l.
  11. The limit of a sum is the sum of the limits, provided each limit exists.
  12. The limit of a product is the product of the limits, provided each limit exists.
  13. The limit of a quotient is the quotient of the limits, provided the denominator limit is non-zero.
  14. For any positive integer n, lim x→a (xⁿ - aⁿ)/(x - a) = n aⁿ⁻¹.
  15. The derivative of a constant function is zero.
  16. The derivative of f(x) = x is 1.

MCQs — 10 Questions with Answers

Q1. If f(x) = x², what is lim[x→0] f(x)?

  • A. 0 ✓
  • B. 1
  • C. Undefined
  • D. ∞

Answer: A — As x approaches 0, x² approaches 0, so the limit is 0 regardless of the function value at x = 0.

Q2. For a body dropped from a cliff with s = 4.9t², the average velocity between t = 1 and t = 2 seconds is:

  • A. 4.9 m/s
  • B. 9.8 m/s
  • C. 14.7 m/s ✓
  • D. 19.6 m/s

Answer: C — Average velocity = (s(2) − s(1))/(2 − 1) = (19.6 − 4.9)/1 = 14.7 m/s.

Q3. Consider f(x) = {1 if x ≤ 0; 2 if x > 0}. Which statement is true?

  • A. lim[x→0⁻] f(x) = 2
  • B. lim[x→0⁺] f(x) = 1
  • C. lim[x→0] f(x) exists
  • D. lim[x→0⁻] f(x) = 1 and lim[x→0⁺] f(x) = 2 ✓

Answer: D — From the left (x < 0), f(x) = 1, so lim[x→0⁻] f(x) = 1; from the right (x > 0), f(x) = 2, so lim[x→0⁺] f(x) = 2.

Q4. For h(x) = (x² − 4)/(x − 2) with x ≠ 2, what is lim[x→2] h(x)?

  • A. 2
  • B. 4 ✓
  • C. Undefined
  • D. 0

Answer: B — Factor: (x² − 4)/(x − 2) = (x + 2)(x − 2)/(x − 2) = x + 2. As x → 2, x + 2 → 4.

Q5. The instantaneous velocity of a body at t = 2 is best understood as:

  • A. Average velocity over 2 seconds
  • B. Distance travelled at exactly t = 2
  • C. The limit of average velocity as the time interval approaches zero ✓
  • D. The slope of the secant line between t = 0 and t = 2

Answer: C — Instantaneous velocity is defined as lim[Δt→0] (Δs/Δt), the limiting value of average velocities over shrinking intervals.

Q6. Which is NOT a correct statement about limits?

  • A. The limit of f(x) as x → a depends only on values near a, not on f(a)
  • B. If left and right limits differ, the limit does not exist
  • C. The limit always equals the function value f(a) ✓
  • D. A limit describes expected behaviour as x approaches a

Answer: C — The limit may differ from f(a); for example, g(x) = |x| (x ≠ 0) has lim[x→0] g(x) = 0 but g(0) is undefined.

Q7. From Table 12.2, as t₁ approaches 2 from the left, the average velocities approach which value?

  • A. 9.8 m/s
  • B. 14.7 m/s
  • C. 19.6 m/s
  • D. Approximately 19.6 m/s (between 19.551 and 19.649) ✓

Answer: D — The table shows average velocities increasing as t₁ → 2⁻ (from 9.8 to 19.551 at t₁ = 1.99), suggesting the instantaneous velocity is approximately 19.6 m/s.

Q8. The slope of the tangent line to a curve y = f(x) at point (a, f(a)) represents:

  • A. The average rate of change of f
  • B. The derivative (instantaneous rate of change) of f at x = a ✓
  • C. The distance travelled by the function
  • D. The limit of the function as x → a

Answer: B — The tangent line's slope is the geometric interpretation of the derivative, which is the instantaneous rate of change at that point.

Q9. If the average velocity between t = 2 and t = 2.01 is 19.649 m/s, and the average velocity between t = 1.99 and t = 2 is 19.551 m/s, which conclusion is valid?

  • A. The instantaneous velocity at t = 2 is exactly 19.6 m/s
  • B. The instantaneous velocity at t = 2 lies between 19.551 and 19.649 m/s ✓
  • C. The right-hand limit and left-hand limit of velocity are different
  • D. The function s = 4.9t² is discontinuous at t = 2

Answer: B — The left-side average (19.551) and right-side average (19.649) bracket the instantaneous velocity, which should lie between them as intervals shrink.

Q10. For lim[x→a] f(x) to equal L, which statement is logically necessary and sufficient?

  • A. f(a) = L
  • B. Both lim[x→a⁻] f(x) = L and lim[x→a⁺] f(x) = L ✓
  • C. f is continuous at x = a
  • D. f is defined in some interval around a

Answer: B — A limit equals L if and only if both the left-hand and right-hand limits equal L; f(a) need not be defined or equal L.

Flashcards

What is the intuitive meaning of the derivative of a function at a point?

The derivative is the instantaneous rate of change of the function, represented geometrically as the slope of the tangent line to the curve at that point.

Define the limit of a function f(x) as x approaches a.

The limit lim[x→a] f(x) = L means f(x) gets arbitrarily close to L as x gets arbitrarily close to a (but x ≠ a).

What is the relationship between average velocity and the slope of a secant line?

Average velocity between two time points equals the rise-over-run (Δs/Δt) of the secant line connecting those points on the distance-time graph.

When does the limit of a function at a point NOT exist?

The limit does not exist when the left-hand limit and right-hand limit are different, even if the function is defined at that point.

What is the left-hand limit of f(x) as x approaches a?

The left-hand limit lim[x→a⁻] f(x) is the expected value of f(x) based on values of f approaching a from the left (x < a).

What is the right-hand limit of f(x) as x approaches a?

The right-hand limit lim[x→a⁺] f(x) is the expected value of f(x) based on values of f approaching a from the right (x > a).

In the cliff-drop example, what does the distance function s = 4.9t² represent?

It represents the distance (in metres) travelled by a dropped body as a function of time t (in seconds), following the physics of free fall.

How does making the time interval h smaller improve the approximation of instantaneous velocity?

As the interval h → 0, the average velocity over [t, t+h] approaches the exact instantaneous velocity at time t, assuming no dramatic change occurs in that tiny interval.

What geometric insight connects the limiting process to the tangent line?

As the secant line interval shrinks, the secant line rotates and approaches the tangent line, whose slope is the derivative (instantaneous rate of change).

For the function g(x) = |x| with x ≠ 0, why is the limit at x = 0 equal to 0 even though g(0) is undefined?

Because lim[x→0] g(x) = 0 depends only on the behaviour of g(x) near x = 0, not on the value of g at x = 0 itself.

Important Board Questions

State what is meant by the left-hand limit of a function f(x) at x = a. Give an example where the left-hand limit and right-hand limit are different. [2 marks]

Define lim[x→a⁻] f(x) as the expected value of f based on values with x < a. Use the piecewise function f(x) = {1 if x ≤ 0; 2 if x > 0} from the text as your example.

For a body dropped from a cliff, s = 4.9t². Using the definition of average velocity, calculate the average velocity between t = 1.5 and t = 2 seconds. Then explain how average velocities over smaller and smaller intervals ending at t = 2 lead to the concept of instantaneous velocity. [5 marks]

Calculate Δs/Δt = [4.9(2²) − 4.9(1.5²)]/(2 − 1.5). Then discuss how shrinking the interval makes the average velocity approach a limit, which is the instantaneous velocity; relate this to the tangent line slope.

Derive that the derivative (instantaneous rate of change) of s = 4.9t² at t = 2 is the limit of the sequence of average velocities as the time interval approaches zero. Using the geometry of secant lines and tangent lines, explain why this limiting process gives the slope of the tangent to the curve at t = 2. [6 marks]

Set up the limit lim[h→0] [4.9(2+h)² − 4.9(4)]/h; show algebraically that this equals 19.6. Then interpret geometrically: secant lines over intervals [2, 2+h] have slopes that approach the tangent slope as h → 0, visualizing the sequence of secant lines rotating toward the tangent.

Assertion-Reason Questions for Limits and Derivatives

Q1. Assertion (A):

The instantaneous velocity of a falling body at time t = 2 seconds equals the slope of the tangent to the distance-time curve at that point.

Reason (R):

As time intervals ending at t = 2 become smaller, the sequence of average velocities approaches the slope of the tangent line to the curve at t = 2.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both statements are true, and the reason correctly explains why the assertion is valid. The limiting process of shrinking intervals leads to the tangent slope, which defines instantaneous velocity.

Q2. Assertion (A):

For the function f(x) = |x| where x ≠ 0, the limit lim(x→0) f(x) = 0 even though f(0) is undefined.

Reason (R):

The existence of a limit at a point depends only on the behavior of the function near that point, not on whether the function is defined at that point.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains the assertion. The limit can exist even if the function is undefined at the point; we only need the function to be defined in a neighborhood around the point.

Q3. Assertion (A):

For a piecewise function where the left-hand limit and right-hand limit at x = a are different, the limit lim(x→a) f(x) does not exist.

Reason (R):

The limit exists if and only if the left-hand limit equals f(a).

Answer: C — A is true but R is false — The assertion is true; if left and right limits differ, the overall limit doesn't exist. However, the reason is false because the correct condition is that left-hand and right-hand limits must be equal to each other—not necessarily equal to f(a).

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Case-Based Questions (CBQ) for Limits and Derivatives

Case Study:

A smartphone manufacturer tests the battery discharge rate of their flagship model. The battery charge Q(t) (in percentage) as a function of time t (in hours) from full charge is modeled by Q(t) = 100 - 2t - 0.1t². An engineer wants to estimate the instantaneous rate of charge loss at exactly t = 5 hours. To do this, she computes the average discharge rate (change in charge per hour) over time intervals ending at t = 5 seconds: from t = 4.5 to 5, from t = 4.9 to 5, from t = 4.99 to 5, and from t = 4.999 to 5.

Q1. What is the battery charge Q(5) at t = 5 hours?

  • A. 80%
  • B. 79.5% ✓
  • C. 78.5%
  • D. 75%

Answer: B

Q2. What is the average discharge rate between t = 4.9 and t = 5 hours?

  • A. 2.09 percentage points per hour ✓
  • B. 2.5 percentage points per hour
  • C. 2.19 percentage points per hour
  • D. 1.9 percentage points per hour

Answer: A

Q3. Based on the limiting process described, what does the engineer expect the instantaneous discharge rate to approach as time intervals become infinitesimally small?

  • A. Approximately 2.0 percentage points per hour
  • B. Approximately 2.5 percentage points per hour
  • C. Approximately 2.1 percentage points per hour ✓
  • D. Approximately 3.0 percentage points per hour

Answer: C

Q4. Which concept from the chapter does this scenario directly illustrate?

  • A. The definition of average rate of change and how it leads to instantaneous rate of change via limits ✓
  • B. The concept of left-hand and right-hand limits being different
  • C. The calculation of polynomial functions
  • D. The slope of a straight line passing through two points

Answer: A

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