High School Mathematics — Final Year

Grade 12: Precalculus

Master advanced functions, trigonometric identities, conic sections, sequences, vectors, and limits — the complete bridge to college-level calculus

Practice Worksheets Available

Master Precalculus concepts with our advanced worksheet collection

Welcome to Precalculus!

This is the most important course before calculus. You'll master advanced functions, trigonometric identities, conic sections, sequences and series, vectors, parametric equations, and an introduction to limits. Every lesson includes detailed concept explanations, worked examples, and 30 practice questions with complete step-by-step solutions.

— Mr. Augustine, Mathematics Department

8
Total Lessons
240+
Practice Questions
5
State Tests
8
Topics Covered

Functions: Review & Advanced Concepts

Master function notation, composition, inverses, and transformations at the precalculus level

Key Vocabulary

FunctionDomainRangeCompositionInverse FunctionOne-to-OneOntoBijectionTransformationPiecewise Function

Concept Explanation

A function assigns exactly one output to each input. Precalculus deepens your understanding of functions as the foundation for calculus. **Function Notation:** f(x) reads "f of x" — x is the input, f(x) is the output. f(a + h) means substitute (a + h) everywhere you see x. **Composition of Functions:** (f ∘ g)(x) = f(g(x)) — apply g first, then f. Domain of f ∘ g: all x in domain of g where g(x) is in domain of f. **Inverse Functions:** f⁻¹ is the inverse of f if f(f⁻¹(x)) = x and f⁻¹(f(x)) = x. To find f⁻¹: swap x and y, then solve for y. A function has an inverse only if it is one-to-one (passes horizontal line test). **Transformations Summary:** • f(x) + k → shift up k • f(x) - k → shift down k • f(x + h) → shift left h • f(x - h) → shift right h • -f(x) → reflect over x-axis • f(-x) → reflect over y-axis • af(x) → vertical stretch/compression by factor a • f(bx) → horizontal compression/stretch by factor 1/b **Piecewise Functions:** Defined by different formulas on different intervals. Always check which piece applies for a given x-value. **Difference Quotient:** [f(x + h) - f(x)] / h — the foundation of the derivative in calculus.

Worked Examples

1

If f(x) = 2x² - 3x + 1 and g(x) = x + 4, find (f ∘ g)(x) and (g ∘ f)(x)

Step 1: Find (f ∘ g)(x) = f(g(x)) Substitute g(x) = x + 4 into f: f(x + 4) = 2(x + 4)² - 3(x + 4) + 1 = 2(x² + 8x + 16) - 3x - 12 + 1 = 2x² + 16x + 32 - 3x - 12 + 1 = 2x² + 13x + 21 Step 2: Find (g ∘ f)(x) = g(f(x)) Substitute f(x) into g: g(2x² - 3x + 1) = (2x² - 3x + 1) + 4 = 2x² - 3x + 5 Answer: (f ∘ g)(x) = 2x² + 13x + 21 and (g ∘ f)(x) = 2x² - 3x + 5
2

Find the inverse of f(x) = (3x - 2)/(x + 1)

Step 1: Replace f(x) with y y = (3x - 2)/(x + 1) Step 2: Swap x and y x = (3y - 2)/(y + 1) Step 3: Solve for y x(y + 1) = 3y - 2 xy + x = 3y - 2 xy - 3y = -2 - x y(x - 3) = -x - 2 y = (-x - 2)/(x - 3) = (x + 2)/(3 - x) Answer: f⁻¹(x) = (x + 2)/(3 - x), domain: x ≠ 3
3

Compute the difference quotient for f(x) = x² + 3x

Step 1: Find f(x + h) f(x + h) = (x + h)² + 3(x + h) = x² + 2xh + h² + 3x + 3h Step 2: Compute f(x + h) - f(x) = x² + 2xh + h² + 3x + 3h - (x² + 3x) = 2xh + h² + 3h Step 3: Divide by h [f(x + h) - f(x)] / h = (2xh + h² + 3h) / h = 2x + h + 3 Answer: Difference quotient = 2x + h + 3

Mr. Augustine's Teaching Tip

The difference quotient [f(x+h) - f(x)]/h is the single most important formula in precalculus — it becomes the derivative in calculus! Practice it with every function type. For inverses, always verify by checking that f(f⁻¹(x)) = x. And remember: composition is NOT commutative — f∘g ≠ g∘f in general!

Polynomial & Rational Functions (Advanced)

Analyze complex polynomials, rational functions, partial fractions, and real-world modeling

Key Vocabulary

Rational Root TheoremDescartes' Rule of SignsFundamental Theorem of AlgebraPartial FractionsOblique AsymptoteMultiplicityConjugate PairsBounds

Concept Explanation

**Rational Root Theorem:** If f(x) = aₙxⁿ + ... + a₀ has integer coefficients, then any rational zero p/q satisfies: • p divides the constant term a₀ • q divides the leading coefficient aₙ **Descartes' Rule of Signs:** • Number of positive real zeros = number of sign changes in f(x), or less by an even number • Number of negative real zeros = number of sign changes in f(-x), or less by an even number **Fundamental Theorem of Algebra:** Every polynomial of degree n ≥ 1 has exactly n zeros (counting multiplicity) in the complex number system. **Complex Conjugate Pairs:** If a + bi is a zero of a polynomial with real coefficients, then a - bi is also a zero. **Partial Fraction Decomposition:** Used to break rational expressions into simpler fractions for integration. For distinct linear factors: A/(x-a) + B/(x-b) For repeated factors: A/(x-a) + B/(x-a)² For irreducible quadratics: (Ax+B)/(x²+bx+c) **Upper and Lower Bounds:** Synthetic division with positive c: if all values in bottom row are non-negative → c is upper bound. Synthetic division with negative c: if values alternate in sign → c is lower bound.

Worked Examples

1

Find all rational zeros of f(x) = 2x³ - 3x² - 11x + 6

Step 1: List possible rational zeros using Rational Root Theorem p divides 6: ±1, ±2, ±3, ±6 q divides 2: ±1, ±2 Possible: ±1, ±2, ±3, ±6, ±1/2, ±3/2 Step 2: Test x = 3 using synthetic division 3 | 2 -3 -11 6 | 6 9 -6 ________________ 2 3 -2 0 ← remainder 0, so x = 3 is a zero Step 3: Factor the quotient 2x² + 3x - 2 2x² + 3x - 2 = (2x - 1)(x + 2) Zeros: x = 1/2 and x = -2 Answer: x = 3, x = 1/2, x = -2
2

Decompose into partial fractions: (3x + 5)/[(x+1)(x-2)]

Step 1: Set up partial fractions (3x + 5)/[(x+1)(x-2)] = A/(x+1) + B/(x-2) Step 2: Multiply both sides by (x+1)(x-2) 3x + 5 = A(x-2) + B(x+1) Step 3: Solve for A and B Let x = 2: 3(2) + 5 = B(3) → 11 = 3B → B = 11/3 Let x = -1: 3(-1) + 5 = A(-3) → 2 = -3A → A = -2/3 Answer: -2/[3(x+1)] + 11/[3(x-2)]
3

Use Descartes' Rule to determine possible positive/negative zeros of f(x) = x⁴ - 3x³ + x² + 2x - 5

Step 1: Count sign changes in f(x) f(x) = x⁴ - 3x³ + x² + 2x - 5 Signs: + - + + - Changes: +→- (1), -→+ (2), +→+ (none), +→- (3) 3 sign changes → 3 or 1 positive real zeros Step 2: Count sign changes in f(-x) f(-x) = x⁴ + 3x³ + x² - 2x - 5 Signs: + + + - - Changes: +→- (1) 1 sign change → exactly 1 negative real zero Answer: 3 or 1 positive zeros; exactly 1 negative zero

Mr. Augustine's Teaching Tip

The Rational Root Theorem gives you a finite list to test — don't try random numbers! Once you find one zero, use synthetic division to reduce the degree, then factor or use the quadratic formula on the quotient. For partial fractions, the 'cover-up' method is fastest for distinct linear factors: cover the factor in the denominator and substitute its zero into the rest.

Exponential & Logarithmic Functions (Advanced)

Master advanced exponential models, logarithmic equations, natural log, and real-world applications

Key Vocabulary

Natural LogarithmEuler's Number eContinuous CompoundingLogistic GrowthExponential RegressionHalf-LifeDoubling TimepH ScaleRichter ScaleDecibels

Concept Explanation

**The Number e:** e ≈ 2.71828... is the base of natural logarithms. e = lim(n→∞) (1 + 1/n)ⁿ Continuous compounding: A = Pe^(rt) **Natural Logarithm:** ln(x) = log_e(x) ln(e) = 1, ln(1) = 0, ln(e^x) = x, e^(ln x) = x **Solving Exponential Equations:** 1. Same base: set exponents equal 2. Different bases: take ln or log of both sides 3. Use power rule: ln(a^x) = x·ln(a) **Solving Logarithmic Equations:** 1. Condense to single log 2. Convert to exponential form 3. Solve and CHECK (domain: argument must be positive) **Exponential Growth/Decay Models:** • Growth: A(t) = A₀e^(kt), k > 0 • Decay: A(t) = A₀e^(-kt), k > 0 • Half-life: t₁/₂ = ln(2)/k • Doubling time: t_d = ln(2)/k **Logistic Growth:** P(t) = L / (1 + Ce^(-kt)) where L = carrying capacity **Real-World Logarithm Applications:** • pH = -log[H⁺] • Richter scale: M = log(I/I₀) • Decibels: dB = 10·log(I/I₀) • Compound interest: A = P(1 + r/n)^(nt)

Worked Examples

1

Solve: 3^(2x-1) = 27^(x+2)

Step 1: Express both sides with base 3 27 = 3³, so 27^(x+2) = 3^(3(x+2)) = 3^(3x+6) Step 2: Set exponents equal 2x - 1 = 3x + 6 Step 3: Solve -1 - 6 = 3x - 2x -7 = x Answer: x = -7
2

Solve: ln(x + 3) + ln(x - 1) = ln(5)

Step 1: Condense left side using product rule ln[(x + 3)(x - 1)] = ln(5) Step 2: Since ln is one-to-one, set arguments equal (x + 3)(x - 1) = 5 x² + 2x - 3 = 5 x² + 2x - 8 = 0 (x + 4)(x - 2) = 0 x = -4 or x = 2 Step 3: CHECK domain (arguments must be positive) x = -4: ln(-4 + 3) = ln(-1) — undefined! Extraneous. x = 2: ln(5) + ln(1) = ln(5) ✓ Answer: x = 2
3

A radioactive substance has a half-life of 10 years. How long until only 25% remains?

Step 1: Set up the decay model A(t) = A₀(1/2)^(t/10) Step 2: Set A(t) = 0.25A₀ 0.25A₀ = A₀(1/2)^(t/10) 0.25 = (1/2)^(t/10) (1/4) = (1/2)^(t/10) (1/2)² = (1/2)^(t/10) Step 3: Set exponents equal 2 = t/10 t = 20 years Answer: 20 years (two half-lives)

Mr. Augustine's Teaching Tip

Always CHECK logarithmic equation solutions — you can only take the log of positive numbers, so extraneous solutions are common! For exponential equations, if you can't match bases, take ln of both sides and use the power rule. Remember: ln and e are inverses, so they cancel each other. For real-world problems, identify whether it's growth (k > 0) or decay (k < 0) first.

Trigonometry: Advanced Identities & Equations

Master sum/difference formulas, double/half-angle identities, and solving trig equations

Key Vocabulary

Sum FormulaDifference FormulaDouble-Angle FormulaHalf-Angle FormulaProduct-to-SumSum-to-ProductGeneral SolutionPrincipal ValueCofunction Identity

Concept Explanation

**Pythagorean Identities:** sin²θ + cos²θ = 1 1 + tan²θ = sec²θ 1 + cot²θ = csc²θ **Sum & Difference Formulas:** sin(A ± B) = sinA cosB ± cosA sinB cos(A ± B) = cosA cosB ∓ sinA sinB tan(A ± B) = (tanA ± tanB) / (1 ∓ tanA tanB) **Double-Angle Formulas:** sin(2θ) = 2sinθ cosθ cos(2θ) = cos²θ - sin²θ = 2cos²θ - 1 = 1 - 2sin²θ tan(2θ) = 2tanθ / (1 - tan²θ) **Half-Angle Formulas:** sin(θ/2) = ±√[(1 - cosθ)/2] cos(θ/2) = ±√[(1 + cosθ)/2] tan(θ/2) = sinθ/(1 + cosθ) = (1 - cosθ)/sinθ **Solving Trig Equations:** 1. Isolate the trig function 2. Find the reference angle 3. Determine all solutions in [0, 2π) 4. Add 2πk (or πk for tan/cot) for general solution **Cofunction Identities:** sin(π/2 - θ) = cosθ cos(π/2 - θ) = sinθ tan(π/2 - θ) = cotθ

Worked Examples

1

Find the exact value of sin(75°)

Step 1: Write 75° as a sum of known angles 75° = 45° + 30° Step 2: Apply sum formula sin(45° + 30°) = sin45°cos30° + cos45°sin30° Step 3: Substitute exact values = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 Answer: sin(75°) = (√6 + √2)/4
2

If sinθ = 3/5 and θ is in QI, find sin(2θ) and cos(2θ)

Step 1: Find cosθ using Pythagorean identity sin²θ + cos²θ = 1 (3/5)² + cos²θ = 1 cos²θ = 1 - 9/25 = 16/25 cosθ = 4/5 (positive in QI) Step 2: Apply double-angle formulas sin(2θ) = 2sinθcosθ = 2(3/5)(4/5) = 24/25 cos(2θ) = cos²θ - sin²θ = 16/25 - 9/25 = 7/25 Answer: sin(2θ) = 24/25, cos(2θ) = 7/25
3

Solve: 2sin²x - sinx - 1 = 0 on [0, 2π)

Step 1: Factor (treat sinx as a variable u) Let u = sinx: 2u² - u - 1 = 0 (2u + 1)(u - 1) = 0 u = -1/2 or u = 1 Step 2: Solve sinx = -1/2 Reference angle: π/6 sinx is negative in QIII and QIV x = π + π/6 = 7π/6 and x = 2π - π/6 = 11π/6 Step 3: Solve sinx = 1 x = π/2 Answer: x = π/2, 7π/6, 11π/6

Mr. Augustine's Teaching Tip

Memorize the sum formulas — everything else derives from them! For double-angle, just use sum formula with A = B = θ. When solving trig equations, always find ALL solutions in [0, 2π) first, then add the period for the general solution. Factor trig equations just like algebraic ones — substitute u = sinx or u = cosx to see the structure clearly.

Conic Sections

Explore parabolas, ellipses, hyperbolas, and circles with equations, graphs, and applications

Key Vocabulary

Conic SectionFocusDirectrixVertexMajor AxisMinor AxisCenterEccentricityAsymptoteLatus RectumTransverse Axis

Concept Explanation

**Circles:** Standard form: (x - h)² + (y - k)² = r² Center: (h, k), Radius: r **Parabolas:** Vertical: (x - h)² = 4p(y - k) Horizontal: (y - k)² = 4p(x - h) • Vertex: (h, k) • Focus: p units from vertex along axis • Directrix: p units from vertex, opposite focus • If p > 0: opens up/right; if p < 0: opens down/left **Ellipses:** Horizontal major axis: (x-h)²/a² + (y-k)²/b² = 1, a > b Vertical major axis: (x-h)²/b² + (y-k)²/a² = 1, a > b • Center: (h, k) • a = semi-major axis, b = semi-minor axis • c² = a² - b² (c = distance from center to focus) • Eccentricity: e = c/a (0 < e < 1) **Hyperbolas:** Horizontal: (x-h)²/a² - (y-k)²/b² = 1 Vertical: (y-k)²/a² - (x-h)²/b² = 1 • c² = a² + b² • Asymptotes: y - k = ±(b/a)(x - h) [horizontal] • Eccentricity: e = c/a (e > 1) **Identifying Conics from General Form Ax² + Bxy + Cy² + Dx + Ey + F = 0:** • Circle: A = C (and B = 0) • Parabola: A = 0 or C = 0 (but not both) • Ellipse: A and C have same sign, A ≠ C • Hyperbola: A and C have opposite signs

Worked Examples

1

Find the equation of the ellipse with vertices (±5, 0) and co-vertices (0, ±3)

Step 1: Identify the center and axes Center: (0, 0) Horizontal major axis (vertices on x-axis) a = 5 (semi-major), b = 3 (semi-minor) Step 2: Write standard form x²/a² + y²/b² = 1 x²/25 + y²/9 = 1 Step 3: Find foci c² = a² - b² = 25 - 9 = 16 c = 4 Foci: (±4, 0) Answer: x²/25 + y²/9 = 1, foci at (±4, 0)
2

Find the vertex, focus, and directrix of x² = 12y

Step 1: Match to standard form (x-h)² = 4p(y-k) x² = 12y → 4p = 12 → p = 3 Vertex: (0, 0) Step 2: Find focus Focus is p = 3 units above vertex (p > 0, opens up) Focus: (0, 3) Step 3: Find directrix Directrix is p = 3 units below vertex Directrix: y = -3 Answer: Vertex (0,0), Focus (0,3), Directrix y = -3
3

Find the asymptotes of the hyperbola x²/9 - y²/16 = 1

Step 1: Identify a and b a² = 9 → a = 3 b² = 16 → b = 4 Center: (0, 0), horizontal transverse axis Step 2: Write asymptote equations y = ±(b/a)x = ±(4/3)x Step 3: Find foci c² = a² + b² = 9 + 16 = 25 → c = 5 Foci: (±5, 0) Answer: Asymptotes y = ±(4/3)x, Foci (±5, 0)

Mr. Augustine's Teaching Tip

The key to conic sections is recognizing which form you have. Always complete the square to convert to standard form — it reveals the center, vertices, and foci immediately. Remember: for ellipses c² = a² - b² (c is smaller than a), but for hyperbolas c² = a² + b² (c is larger than a). The eccentricity tells you the shape: e = 0 is a circle, 0 < e < 1 is an ellipse, e = 1 is a parabola, e > 1 is a hyperbola.

Sequences, Series & Mathematical Induction

Master arithmetic and geometric sequences, series, sigma notation, and proof by induction

Key Vocabulary

Arithmetic SequenceGeometric SequenceCommon DifferenceCommon RatioPartial SumInfinite SeriesConvergenceSigma NotationMathematical InductionBinomial TheoremPascal's Triangle

Concept Explanation

**Arithmetic Sequences:** aₙ = a₁ + (n-1)d, where d = common difference Sum of n terms: Sₙ = n/2 · (a₁ + aₙ) = n/2 · [2a₁ + (n-1)d] **Geometric Sequences:** aₙ = a₁ · r^(n-1), where r = common ratio Sum of n terms: Sₙ = a₁(1 - rⁿ)/(1 - r), r ≠ 1 **Infinite Geometric Series:** S = a₁/(1 - r), only if |r| < 1 (converges) If |r| ≥ 1, the series diverges. **Sigma Notation:** Σᵢ₌₁ⁿ aᵢ = a₁ + a₂ + ... + aₙ **Useful Summation Formulas:** Σᵢ₌₁ⁿ 1 = n Σᵢ₌₁ⁿ i = n(n+1)/2 Σᵢ₌₁ⁿ i² = n(n+1)(2n+1)/6 Σᵢ₌₁ⁿ i³ = [n(n+1)/2]² **Mathematical Induction:** To prove P(n) for all n ≥ 1: 1. Base case: Verify P(1) is true 2. Inductive step: Assume P(k) is true, prove P(k+1) is true **Binomial Theorem:** (a + b)ⁿ = Σₖ₌₀ⁿ C(n,k) · aⁿ⁻ᵏ · bᵏ where C(n,k) = n! / [k!(n-k)!]

Worked Examples

1

Find the 20th term and sum of the first 20 terms of the arithmetic sequence 3, 7, 11, 15, ...

Step 1: Identify a₁ and d a₁ = 3, d = 7 - 3 = 4 Step 2: Find a₂₀ a₂₀ = a₁ + (20-1)d = 3 + 19(4) = 3 + 76 = 79 Step 3: Find S₂₀ S₂₀ = 20/2 · (a₁ + a₂₀) = 10 · (3 + 79) = 10 · 82 = 820 Answer: a₂₀ = 79, S₂₀ = 820
2

Find the sum of the infinite geometric series: 8 + 4 + 2 + 1 + ...

Step 1: Identify a₁ and r a₁ = 8, r = 4/8 = 1/2 Step 2: Check convergence |r| = 1/2 < 1 → converges Step 3: Apply infinite sum formula S = a₁/(1 - r) = 8/(1 - 1/2) = 8/(1/2) = 16 Answer: S = 16
3

Expand (x + 2)⁴ using the Binomial Theorem

Step 1: Apply Binomial Theorem (x + 2)⁴ = Σₖ₌₀⁴ C(4,k) · x^(4-k) · 2^k Step 2: Calculate each term k=0: C(4,0)·x⁴·1 = x⁴ k=1: C(4,1)·x³·2 = 4·2x³ = 8x³ k=2: C(4,2)·x²·4 = 6·4x² = 24x² k=3: C(4,3)·x·8 = 4·8x = 32x k=4: C(4,4)·1·16 = 16 Answer: x⁴ + 8x³ + 24x² + 32x + 16

Mr. Augustine's Teaching Tip

For arithmetic sequences, the common difference d is constant — check by subtracting consecutive terms. For geometric sequences, the common ratio r is constant — check by dividing consecutive terms. The infinite geometric series formula S = a₁/(1-r) only works when |r| < 1! For the Binomial Theorem, Pascal's Triangle gives you the coefficients C(n,k) quickly for small n.

Vectors & Parametric Equations

Master vector operations, dot products, parametric curves, and applications in physics and geometry

Key Vocabulary

VectorScalarMagnitudeDirection AngleUnit VectorDot ProductComponent FormResultantParametric EquationsParameterEliminating the Parameter

Concept Explanation

**Vectors:** A vector has both magnitude and direction. Component form: v = ⟨a, b⟩ Magnitude: |v| = √(a² + b²) Direction angle: θ = arctan(b/a) **Vector Operations:** • Addition: ⟨a,b⟩ + ⟨c,d⟩ = ⟨a+c, b+d⟩ • Subtraction: ⟨a,b⟩ - ⟨c,d⟩ = ⟨a-c, b-d⟩ • Scalar multiplication: k⟨a,b⟩ = ⟨ka, kb⟩ • Unit vector: û = v/|v| **Dot Product:** u · v = a₁a₂ + b₁b₂ |u||v|cosθ = u · v • If u · v = 0, vectors are perpendicular • If u · v > 0, angle is acute • If u · v < 0, angle is obtuse **Angle Between Vectors:** cosθ = (u · v) / (|u| · |v|) **Parametric Equations:** Express x and y as functions of a parameter t: x = f(t), y = g(t) To eliminate the parameter: solve one equation for t, substitute into the other. **Common Parametric Curves:** • Line: x = x₀ + at, y = y₀ + bt • Circle: x = r·cos(t), y = r·sin(t) • Ellipse: x = a·cos(t), y = b·sin(t) **Projectile Motion:** x = v₀cos(θ)·t y = v₀sin(θ)·t - (1/2)gt²

Worked Examples

1

Find the magnitude and direction angle of v = ⟨3, -4⟩

Step 1: Find magnitude |v| = √(3² + (-4)²) = √(9 + 16) = √25 = 5 Step 2: Find direction angle θ = arctan(-4/3) ≈ -53.1° Since the vector is in QIV (positive x, negative y): θ ≈ -53.1° or equivalently 306.9° Answer: |v| = 5, θ ≈ -53.1° (or 306.9°)
2

Find the angle between u = ⟨2, 1⟩ and v = ⟨1, 3⟩

Step 1: Compute dot product u · v = (2)(1) + (1)(3) = 2 + 3 = 5 Step 2: Compute magnitudes |u| = √(4 + 1) = √5 |v| = √(1 + 9) = √10 Step 3: Apply angle formula cosθ = (u · v)/(|u||v|) = 5/(√5 · √10) = 5/√50 = 5/(5√2) = 1/√2 θ = arccos(1/√2) = 45° Answer: θ = 45°
3

Eliminate the parameter: x = t + 1, y = t² - 2

Step 1: Solve for t from the x equation x = t + 1 → t = x - 1 Step 2: Substitute into y equation y = t² - 2 = (x - 1)² - 2 y = x² - 2x + 1 - 2 y = x² - 2x - 1 Answer: y = x² - 2x - 1 (a parabola)

Mr. Augustine's Teaching Tip

Vectors are everywhere in physics and engineering! The dot product is your tool for finding angles and checking perpendicularity. Remember: two vectors are perpendicular if and only if their dot product is zero. For parametric equations, think of t as time — the curve traces out a path as t increases. When eliminating the parameter, always check if there are domain restrictions on t that limit the curve.

Limits & Introduction to Calculus

Understand limits intuitively and formally, continuity, and the bridge to differential calculus

Key Vocabulary

LimitOne-Sided LimitContinuityRemovable DiscontinuityJump DiscontinuityInfinite DiscontinuitySqueeze TheoremIndeterminate FormL'Hôpital's RuleDerivative

Concept Explanation

**The Concept of a Limit:** lim(x→c) f(x) = L means f(x) gets arbitrarily close to L as x approaches c. The limit does NOT depend on f(c) — only on values near c. **One-Sided Limits:** • Left-hand limit: lim(x→c⁻) f(x) • Right-hand limit: lim(x→c⁺) f(x) • Two-sided limit exists ⟺ both one-sided limits exist and are equal **Limit Laws:** • lim[f(x) ± g(x)] = lim f(x) ± lim g(x) • lim[f(x)·g(x)] = lim f(x) · lim g(x) • lim[f(x)/g(x)] = lim f(x) / lim g(x), if lim g(x) ≠ 0 • lim[f(x)]ⁿ = [lim f(x)]ⁿ **Continuity:** f is continuous at x = c if: 1. f(c) is defined 2. lim(x→c) f(x) exists 3. lim(x→c) f(x) = f(c) **Types of Discontinuity:** • Removable: hole in graph (limit exists, but f(c) ≠ limit) • Jump: left and right limits exist but are unequal • Infinite: vertical asymptote **Indeterminate Forms:** 0/0, ∞/∞ — need algebraic manipulation or L'Hôpital's Rule **The Derivative (Preview):** f'(x) = lim(h→0) [f(x+h) - f(x)] / h This is the instantaneous rate of change — the slope of the tangent line. **Special Limits:** lim(x→0) sinx/x = 1 lim(x→0) (1-cosx)/x = 0 lim(x→∞) (1 + 1/x)^x = e

Worked Examples

1

Evaluate: lim(x→3) (x² - 9)/(x - 3)

Step 1: Check for indeterminate form Direct substitution: (9-9)/(3-3) = 0/0 — indeterminate Step 2: Factor and simplify (x² - 9)/(x - 3) = (x-3)(x+3)/(x-3) = x + 3 (for x ≠ 3) Step 3: Take the limit lim(x→3) (x + 3) = 3 + 3 = 6 Answer: 6
2

Determine if f(x) = {x² if x < 2; 4 if x = 2; x + 3 if x > 2} is continuous at x = 2

Step 1: Check f(2) f(2) = 4 ✓ (defined) Step 2: Find left-hand limit lim(x→2⁻) x² = 4 Step 3: Find right-hand limit lim(x→2⁺) (x + 3) = 5 Step 4: Compare Left limit (4) ≠ Right limit (5) The two-sided limit does not exist. Answer: f is NOT continuous at x = 2 (jump discontinuity)
3

Use the limit definition to find f'(x) for f(x) = x²

Step 1: Set up the difference quotient f'(x) = lim(h→0) [f(x+h) - f(x)] / h = lim(h→0) [(x+h)² - x²] / h Step 2: Expand = lim(h→0) [x² + 2xh + h² - x²] / h = lim(h→0) [2xh + h²] / h = lim(h→0) h(2x + h) / h = lim(h→0) (2x + h) Step 3: Evaluate = 2x + 0 = 2x Answer: f'(x) = 2x

Mr. Augustine's Teaching Tip

Limits are the foundation of all calculus! When you get 0/0, don't panic — it just means you need to do more algebra (factor, rationalize, or use trig identities). The limit exists only when BOTH one-sided limits agree. For continuity, think of it as 'you can draw the graph without lifting your pencil.' The derivative definition lim(h→0)[f(x+h)-f(x)]/h is the most important formula in calculus — understand it deeply!

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