Guided Practice
Apply your knowledge of limits and sequences through structured practice problems with detailed step-by-step solutions.
Learn
This guided practice session consolidates your understanding of limits and sequences. Each problem builds upon fundamental concepts, progressing from basic evaluation to more complex applications.
Key Concepts Review
- Limit Definition: The value a function approaches as the input approaches a specific value
- Direct Substitution: When the function is continuous at the point, substitute directly
- Factoring Method: Factor and simplify when direct substitution yields 0/0
- Arithmetic Sequences: a_n = a_1 + (n-1)d where d is the common difference
- Geometric Sequences: a_n = a_1 * r^(n-1) where r is the common ratio
- Series Summation: The sum of terms in a sequence, finite or infinite
Problem-Solving Strategy
- Identify the type of problem (limit evaluation, sequence pattern, series sum)
- Determine the appropriate technique based on the form
- Execute the calculation carefully, showing all steps
- Verify your answer using estimation or alternative methods
Examples
Work through these guided examples to strengthen your problem-solving skills.
Example 1: Evaluating a Limit by Factoring
Problem: Find lim(x->3) [(x^2 - 9)/(x - 3)]
Step 1: Try direct substitution: (9 - 9)/(3 - 3) = 0/0 (indeterminate form)
Step 2: Factor the numerator: x^2 - 9 = (x + 3)(x - 3)
Step 3: Simplify: [(x + 3)(x - 3)]/(x - 3) = x + 3 for x != 3
Step 4: Evaluate the simplified expression: lim(x->3) (x + 3) = 6
Answer: The limit equals 6
Example 2: Finding the Sum of an Arithmetic Series
Problem: Find the sum of the first 20 terms: 5 + 8 + 11 + 14 + ...
Step 1: Identify the sequence type: Arithmetic (d = 3)
Step 2: Find a_20: a_20 = 5 + (20-1)(3) = 5 + 57 = 62
Step 3: Apply the sum formula: S_n = n(a_1 + a_n)/2
Step 4: Calculate: S_20 = 20(5 + 62)/2 = 20(67)/2 = 670
Answer: The sum is 670
Example 3: Infinite Geometric Series
Problem: Find the sum: 12 + 4 + 4/3 + 4/9 + ...
Step 1: Identify r: r = 4/12 = 1/3
Step 2: Since |r| < 1, the series converges
Step 3: Apply the formula: S = a_1/(1 - r) = 12/(1 - 1/3) = 12/(2/3) = 18
Answer: The sum is 18
Practice
Complete these practice problems to reinforce your understanding. Show all work.
Problem 1: Evaluate lim(x->2) [(x^2 - 4)/(x - 2)]
Problem 2: Find lim(x->0) [(3x^2 + 5x)/x]
Problem 3: Evaluate lim(x->4) [(sqrt(x) - 2)/(x - 4)] (Hint: rationalize the numerator)
Problem 4: Find the 15th term of the arithmetic sequence: 7, 12, 17, 22, ...
Problem 5: Find the sum of the first 25 terms of the arithmetic sequence: 3, 7, 11, 15, ...
Problem 6: Find the 8th term of the geometric sequence: 2, 6, 18, 54, ...
Problem 7: Determine if the infinite series converges or diverges. If it converges, find the sum: 27 + 9 + 3 + 1 + ...
Problem 8: Find lim(x->infinity) [(3x^2 + 2x)/(x^2 - 1)]
Problem 9: Write an explicit formula for the sequence: -2, 4, -8, 16, -32, ...
Problem 10: Find the sum of the infinite geometric series: 100 + 20 + 4 + 0.8 + ...
Check Your Understanding
Verify your answers and assess your comprehension.
Answer Key
- lim(x->2) [(x^2 - 4)/(x - 2)] = lim(x->2) (x + 2) = 4
- lim(x->0) [(3x^2 + 5x)/x] = lim(x->0) (3x + 5) = 5
- lim(x->4) [(sqrt(x) - 2)/(x - 4)] = 1/(sqrt(4) + 2) = 1/4
- a_15 = 7 + (15-1)(5) = 7 + 70 = 77
- a_25 = 3 + 24(4) = 99; S_25 = 25(3 + 99)/2 = 1275
- a_8 = 2(3)^7 = 2(2187) = 4374
- r = 1/3, |r| < 1, converges; S = 27/(1 - 1/3) = 27/(2/3) = 40.5
- lim(x->infinity) [(3x^2 + 2x)/(x^2 - 1)] = 3
- a_n = -2(-2)^(n-1) or a_n = (-1)^n * 2^n
- r = 1/5, S = 100/(1 - 1/5) = 100/(4/5) = 125
Self-Assessment
- Score 9-10: Excellent! You've mastered the concepts.
- Score 7-8: Good understanding. Review any missed problems.
- Score 5-6: Revisit the lessons on limits and sequences before proceeding.
- Score below 5: Consider reviewing Lessons 1 and 2 before continuing.
Next Steps
- Review any problems you found challenging
- Practice similar problems from your textbook or additional resources
- Proceed to Word Problems to apply these concepts in context
- Note any recurring difficulties to address in the Common Mistakes lesson