Limit of a function Worksheet for 12th Grade | Lesson Planet - Free Printable

I can see this is a worksheet with limit problems. Let me solve each one step by step.

Looking at the problems, I need to find limits of various functions as x approaches certain values. Let me work through each problem carefully.

Problem 1: lim (x→2) (x² - 4)/(x - 2)
- Factor numerator: (x + 2)(x - 2)/(x - 2)
- Cancel (x - 2): x + 2
- As x → 2: 2 + 2 = 4

Problem 2: lim (x→0) sin(3x)/x
- Use the standard limit: lim (x→0) sin(ax)/x = a
- Answer: 3

Problem 3: lim (x→∞) (x² - 9)/(x² + 3x - 4)
- Divide numerator and denominator by highest power (x²)
- (1 - 9/x²)/(1 + 3/x - 4/x²)
- As x → ∞: (1 - 0)/(1 + 0 - 0) = 1/1 = 1

Problem 4: lim (x→1) (x³ - 1)/(x - 1)
- Factor numerator using difference of cubes: (x - 1)(x² + x + 1)/(x - 1)
- Cancel (x - 1): x² + x + 1
- As x → 1: 1 + 1 + 1 = 3

Problem 5: lim (x→0) [√(x + 4) - 2]/x
- Multiply by conjugate: [√(x + 4) - 2][√(x + 4) + 2]/[x(√(x + 4) + 2)]
- Numerator becomes: (x + 4) - 4 = x
- Result: x/[x(√(x + 4) + 2)] = 1/(√(x + 4) + 2)
- As x → 0: 1/(√4 + 2) = 1/(2 + 2) = 1/4

Problem 6: lim (x→2) (x² - 5x + 6)/(x² - 4)
- Factor: (x - 2)(x - 3)/[(x - 2)(x + 2)]
- Cancel (x - 2): (x - 3)/(x + 2)
- As x → 2: (2 - 3)/(2 + 2) = -1/4

Problem 7: lim (x→∞) (√(x² + 3x))/(x + 1)
- For large x, √(x² + 3x) ≈ √(x²) = |x| = x (since x > 0)
- So we have approximately: x/(x + 1) = 1/(1 + 1/x)
- As x → ∞: 1/(1 + 0) = 1

Problem 8: lim (x→0) (e^x - 1)/x
- This is a standard limit = 1

Problem 9: lim (x→∞) (3x² + 2x - 1)/(x² + 4)
- Divide by x²: (3 + 2/x - 1/x²)/(1 + 4/x²)
- As x → ∞: (3 + 0 - 0)/(1 + 0) = 3

Problem 10: lim (x→0) (1 - cos(x))/x²
- This is a standard limit = 1/2

Problem 11: lim (x→1) (x³ - 1)/(x² - 1)
- Factor: (x - 1)(x² + x + 1)/[(x - 1)(x + 1)]
- Cancel (x - 1): (x² + x + 1)/(x + 1)
- As x → 1: (1 + 1 + 1)/(1 + 1) = 3/2

Problem 12: lim (x→∞) ln(x)/x
- Using L'Hôpital's rule or knowing that polynomial grows faster than logarithm
- Answer: 0

Problem 13: lim (x→0) sin(2x)/sin(3x)
- Rewrite as: [sin(2x)/2x] · [3x/sin(3x)] · (2/3)
- As x → 0: 1 · 1 · (2/3) = 2/3

Problem 14: lim (x→π/2) tan(x)·cos(x)
- tan(x)·cos(x) = sin(x)/cos(x) · cos(x) = sin(x)
- As x → π/2: sin(π/2) = 1

Problem 15: lim (x→∞) (x + √x)/(x - √x)
- Divide by x: (1 + 1/√x)/(1 - 1/√x)
- As x → ∞: (1 + 0)/(1 - 0) = 1

Problem 16: lim (x→0) (1 - cos(x))/x
- This equals 0 (can be shown using Taylor series or L'Hôpital's rule)

Let me verify my answers before providing the final answer.

Final Answer:
1. 4
2. 3
3. 1
4. 3
5. 1/4
6. -1/4
7. 1
8. 1
9. 3
10. 1/2
11. 3/2
12. 0
13. 2/3
14. 1
15. 1
16. 0