Unit 1 Practice Test

Limits and Continuity - 15 Multiple Choice + 6 Free Response Questions

← Back to Unit 1
⏱️ Test Timer
00:00

Recommended time: 45 minutes

📋 Test Instructions

💡 Important: This practice test covers all topics from Unit 1: Limits and Continuity. Take your time and show all your work for free-response questions.
  • Multiple Choice: Select the best answer for each question
  • Free Response: Show all steps and justify your reasoning
  • Time Limit: 45 minutes recommended (AP exam timing)
  • Calculator: Graphing calculator allowed for some questions

📝 Section A: Multiple Choice (15 questions)

1
Find limx→2 (x²-4)/(x-2)
2
Find limx→0 sin(3x)/x
3
Find limx→∞ (2x²+3x+1)/(x²+5)
4
Is f(x) = 1/x continuous at x = 0?
5
Find limx→0 (√(x+4) - 2)/x
6
Find limx→0 (1-cos(x))/x²
7
Find limx→∞ (3x+2)/(2x-1)
8
Find limx→2 |x-2|/(x-2)
9
Find limx→0 (√(x²+9) - 3)/x²
10
Find limx→∞ (1+2/x)ˣ
11
Is f(x) = {x² if x≠2, 5 if x=2} continuous at x=2?
12
Find limx→0 tan(x)/x
13
Find limx→1 ln(x)/(x-1)
14
Find limx→∞ √(x²+3x) - x
15
Find limx→0 (sin(x) - x)/x³

✍️ Section B: Free Response (6 questions)

1
Find limx→1 (x³-1)/(x-1) using factoring.

Solution:

1) Factor the numerator: x³-1 = (x-1)(x²+x+1)

2) Cancel common factor: (x³-1)/(x-1) = (x-1)(x²+x+1)/(x-1) = x²+x+1

3) Take the limit: limx→1 (x²+x+1) = 1²+1+1 = 3

Answer: 3

2
For what value(s) of k is f(x) = {kx+2 if x≤1, x² if x>1} continuous at x=1?

Solution:

1) For continuity at x=1, we need: limx→1⁻ f(x) = limx→1⁺ f(x) = f(1)

2) limx→1⁻ f(x) = limx→1⁻ (kx+2) = k(1)+2 = k+2

3) limx→1⁺ f(x) = limx→1⁺ x² = 1² = 1

4) f(1) = k(1)+2 = k+2 (using the first piece since x≤1)

5) Set equal: k+2 = 1, so k = -1

Answer: k = -1

3
Find limx→0 (eˣ-1)/x using the special limit.

Solution:

1) This is a special limit: limx→0 (eˣ-1)/x = 1

2) No additional work needed - this is a memorized result

3) The limit equals 1 by definition

Answer: 1

4
Find limx→0 (√(x+4) - 2)/x using rationalization.

Solution:

1) Multiply numerator and denominator by conjugate: (√(x+4) + 2)

2) (√(x+4) - 2)(√(x+4) + 2) = (x+4) - 4 = x

3) Denominator becomes: x(√(x+4) + 2)

4) Cancel x: 1/(√(x+4) + 2)

5) Take limit: 1/(√(0+4) + 2) = 1/(2+2) = 1/4

Answer: 1/4

5
Find limx→∞ √(x²+3x) - x using rationalization.

Solution:

1) Multiply by conjugate: (√(x²+3x) + x)/(√(x²+3x) + x)

2) Numerator: (√(x²+3x))² - x² = x²+3x - x² = 3x

3) Denominator: √(x²+3x) + x ≈ √x² + x = x + x = 2x (for large x)

4) Limit becomes: limx→∞ 3x/(2x) = 3/2

Answer: 3/2

6
Find limx→0 (sin(x) - x)/x³ using Taylor series or L'Hôpital's rule.

Solution:

1) Using L'Hôpital's rule (0/0 form):

2) limx→0 (cos(x) - 1)/(3x²) = 0/0

3) Apply again: limx→0 (-sin(x))/(6x) = 0/0

4) Apply once more: limx→0 (-cos(x))/6 = -1/6

Answer: -1/6

📊 Test Results

📚 Study Recommendations

If you scored 80%+:
  • Great job! You have a solid foundation
  • Focus on the few topics you missed
  • Move on to Unit 2: Differentiation
  • Practice more challenging problems
If you scored below 80%:
  • Review the concepts you struggled with
  • Practice more basic limit problems
  • Focus on continuity definitions
  • Use the main Unit 1 page for review