**Definition**: A derivative measures the rate of change of a function at a specific point. It tells us how quickly a function's output changes when its input changes.
**Physical Context - Instantaneous Velocity**: Consider a body falling from a cliff with distance function s = 4.9t² metres (where t is time in seconds). To find the instantaneous velocity at t = 2 seconds:
**Geometric Interpretation**: The instantaneous velocity equals the **slope of the tangent line** to the curve at that point. As time intervals approach zero, the secant line connecting two points on the curve approaches the tangent line at the fixed point.
**Key Insight**: The derivative at a point represents:
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**Definition (Informal)**: We say that the limit of function f(x) as x approaches a number a equals L (written as lim(x→a) f(x) = L) if the values of f(x) get arbitrarily close to L whenever x is sufficiently close to a (but not necessarily equal to a).
**Critical Point**: The limit value L depends ONLY on values of f(x) near a, NOT on whether f(a) exists or what f(a) equals.
**Left-Hand Limit**: lim(x→a⁻) f(x) is the expected value of f(x) as x approaches a from the LEFT (all x values less than a).
**Right-Hand Limit**: lim(x→a⁺) f(x) is the expected value of f(x) as x approaches a from the RIGHT (all x values greater than a).
**Existence of Limit**:
**Example 1**: For f(x) = |x|/x (x ≠ 0):
**Example 2**: For f(x) = x + 10:
**Example 3**: For f(x) = (x² - 4)/(x - 2), x ≠ 2:
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**Theorem 1**: If lim(x→a) f(x) and lim(x→a) g(x) both exist, then:
**(i) Sum Rule**:
lim(x→a) [f(x) + g(x)] = lim(x→a) f(x) + lim(x→a) g(x)
**Proof Idea**: If f(x) → L₁ and g(x) → L₂ as x → a, then f(x) + g(x) → L₁ + L₂
**(ii) Difference Rule**:
lim(x→a) [f(x) - g(x)] = lim(x→a) f(x) - lim(x→a) g(x)
**(iii) Product Rule**:
lim(x→a) [f(x) · g(x)] = lim(x→a) f(x) · lim(x→a) g(x)
**Proof Idea**: If f(x) → L₁ and g(x) → L₂, then products f(x)·g(x) → L₁·L₂
**(iv) Quotient Rule**:
lim(x→a) [f(x)/g(x)] = [lim(x→a) f(x)] / [lim(x→a) g(x)]
**CONDITION**: The denominator limit must be non-zero: lim(x→a) g(x) ≠ 0
**(v) Scalar Multiple Rule** (Special case of product rule):
lim(x→a) [λ·f(x)] = λ · lim(x→a) f(x), where λ is a constant
**Example**: Find lim(x→1) (x² + 3x)
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**Definition**: f(x) = a₀ + a₁x + a₂x² + ... + aₙxⁿ where aᵢ are real numbers and aₙ ≠ 0
**Fundamental Limit**: lim(x→a) x = a
**Power Limits**: By repeated application of product rule:
**Polynomial Limit Theorem**:
lim(x→a) f(x) = f(a)
**Proof**: For polynomial f(x) = a₀ + a₁x + a₂x² + ... + aₙxⁿ:
lim(x→a) f(x) = lim(x→a) [a₀ + a₁x + a₂x² + ... + aₙxⁿ]
= lim(x→a) a₀ + lim(x→a) a₁x + lim(x→a) a₂x² + ... + lim(x→a) aₙxⁿ (sum rule)
= a₀ + a₁·a + a₂·a² + ... + aₙ·aⁿ (using scalar multiple and power limits)
= f(a)
**Example**: lim(x→2) (x³ - 2x² + 5)
= 2³ - 2(2²) + 5 = 8 - 8 + 5 = 5
**Definition**: f(x) = g(x)/h(x) where g(x) and h(x) are polynomials and h(x) ≠ 0
**Case 1 - Non-zero denominator limit** [h(a) ≠ 0]:
lim(x→a) [g(x)/h(x)] = g(a)/h(a) = lim(x→a) g(x) / lim(x→a) h(x)
Simply substitute x = a if h(a) ≠ 0
**Example**: lim(x→1) [(x² + 3)/(x + 2)]
= (1² + 3)/(1 + 2) = 4/3
**Case 2 - Zero denominator, non-zero numerator** [h(a) = 0, g(a) ≠ 0]:
The limit does NOT exist (function becomes unbounded)
**Example**: lim(x→2) [1/(x - 2)]
**Case 3 - Both numerator and denominator zero** [h(a) = 0, g(a) = 0]:
This is an **indeterminate form 0/0**. Use **factorization and cancellation**:
**Example**: lim(x→2) [(x² - 4)/(x - 2)]
**Example**: lim(x→1) [(x³ - 1)/(x² - 1)]
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**Critical Mistakes to Avoid**:
1. **Confusing limit with function value**: lim(x→a) f(x) ≠ f(a) always. The function need not be defined at a for the limit to exist.
2. **Forgetting to check denominator**: When using quotient rule, ALWAYS verify the denominator limit is non-zero.
3. **Not simplifying before substitution**: For 0/0 forms, you MUST factor and cancel before substituting.
4. **Ignoring one-sided limits**: A limit exists ONLY when BOTH left and right limits are equal.
5. **Assuming continuity**: Just because you can draw the graph doesn't mean the limit exists at all points.
**Exam Strategy**:
**Standard Limits to Remember**:
Q1. If f(x) = x², what is lim[x→0] f(x)?
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:
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?
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)?
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:
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?
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?
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:
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?
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?
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.
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.
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.
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