Determining Limits Using Algebraic Manipulation - AP Calculus AB

1. Evaluate $\frac{x^2 - 1}{x - 1}$ as $x$ approaches 1.

   2. Factor: $\frac{(x-1)(x+1)}{x-1} = x+1$, then substitute $x = 1$.

2. State the limit of $f(x) = \frac{x^2 - 9}{x - 3}$ as $x$ approaches 3.

   6. Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, then substitute $x = 3$.

3. Find the limit of $f(x) = x^2 - 4x + 4$ as $x$ approaches 2.

   0. Direct substitution: $2^2 - 4(2) + 4 = 0$.

4. Evaluate the limit of $f(x) = \frac{x^2 - 1}{x - 1}$ as $x$ approaches 1.

   2. Factor: $\frac{(x-1)(x+1)}{x-1} = x+1$, then substitute $x = 1$.

5. Evaluate $\frac{x^2 - 4}{x - 2}$ as $x$ approaches 2.

   4. Factor: $\frac{(x-2)(x+2)}{x-2} = x+2$, then substitute $x = 2$.

6. What is the limit of $\frac{x^2 - 9}{x - 3}$ as $x$ approaches 3?

   6. Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, then substitute $x = 3$.

7. What is the limit of $f(x) = x^2$ as $x$ approaches 3?

   9. Direct substitution: $3^2 = 9$.

8. Determine the limit of $x^2 - 4x + 4$ as $x$ approaches 4.

   4. Direct substitution: $4^2 - 4(4) + 4 = 4$.

9. Evaluate $\frac{2x^2 - 8}{x - 2}$ as $x$ approaches 2.

   8. Factor: $\frac{2(x-2)(x+2)}{x-2} = 2(x+2)$, then substitute $x = 2$.

10. Determine the limit of $f(x) = \frac{x^2 - 1}{x^2 + 1}$ as $x$ approaches infinity.

    1. Divide by highest power: $\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} \to 1$.

11. What is the limit of $h(x) = \frac{x^2 + x - 6}{x - 2}$ as $x$ approaches 2?

    5. Factor: $\frac{(x+3)(x-2)}{x-2} = x+3$, then substitute $x = 2$.

12. Find the limit of $f(x) = x^3$ as $x$ approaches 2.

    8. Direct substitution: $2^3 = 8$.

13. What is the limit of $\frac{x^3 - 8}{x - 2}$ as $x$ approaches 2?

    12. Factor: $\frac{(x-2)(x^2+2x+4)}{x-2} = x^2+2x+4$, then substitute $x = 2$.

14. Evaluate the limit of $\frac{x^2 - 9}{x - 3}$ as $x$ approaches 3.

    6. Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, then substitute $x = 3$.

15. Find the limit of $f(x) = \frac{x^2 - 1}{x^2 + 1}$ as $x$ approaches infinity.

    1. Divide by highest power: $\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} \to 1$.

16. Find the limit of $\frac{x^2 - 25}{x - 5}$ as $x$ approaches 5.

    10. Factor: $\frac{(x-5)(x+5)}{x-5} = x+5$, then substitute $x = 5$.

17. Determine the limit of $f(x) = \frac{1}{x}$ as $x$ approaches infinity.

    0. As $x \to \infty$, $\frac{1}{x} \to 0$.

18. State the limit of $f(x) = \frac{x^2 + 2x + 1}{x + 1}$ as $x$ approaches -1.

    0. Factor: $\frac{(x+1)^2}{x+1} = x+1$, then substitute $x = -1$.

19. State the limit of $f(x) = \frac{1}{x}$ as $x$ approaches negative infinity.

    0. As $x \to -\infty$, $\frac{1}{x} \to 0$.

20. Evaluate $\frac{x^2 - 25}{x - 5}$ as $x$ approaches 5.

    10. Factor: $\frac{(x-5)(x+5)}{x-5} = x+5$, then substitute $x = 5$.

21. What is the limit of $f(x) = \frac{x^2 - 16}{x - 4}$ as $x$ approaches 4?

    8. Factor: $\frac{(x-4)(x+4)}{x-4} = x+4$, then substitute $x = 4$.

22. Determine the limit of $f(x) = \frac{x^2 - 1}{x^2 + 1}$ as $x$ approaches $\infty$.

    1. Divide by highest power: $\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} \to 1$.

23. What is the limit of $g(x) = \frac{1}{x}$ as $x$ approaches 0?

    Does not exist. Function undefined at $x = 0$; left and right limits differ.

24. Find the limit of $f(x) = x^3$ as $x$ approaches -2.

    -8. Direct substitution: $(-2)^3 = -8$.

25. What is the limit of $f(x) = x^3$ as $x$ approaches -1?

    -1. Direct substitution: $(-1)^3 = -1$.

26. Evaluate $\frac{x^2 - 4x + 4}{x - 2}$ as $x$ approaches 2.

    0. Factor: $\frac{(x-2)^2}{x-2} = x-2$, then substitute $x = 2$.

27. Find the limit of $\frac{x^2 - 4x + 4}{x - 2}$ as $x$ approaches 2.

    0. Factor: $\frac{(x-2)^2}{x-2} = x-2$, then substitute $x = 2$.

28. Determine the limit of $f(x) = \frac{x^2 - 1}{x - 1}$ as $x$ approaches 1.

    2. Factor: $\frac{(x-1)(x+1)}{x-1} = x+1$, then substitute $x = 1$.

29. Evaluate the limit of $\frac{x^2 - 4}{x - 2}$ as $x$ approaches 2.

    4. Factor: $\frac{(x-2)(x+2)}{x-2} = x+2$, then substitute $x = 2$.

30. What is the limit of $f(x) = x^3$ as $x$ approaches 3?

    27. Direct substitution: $3^3 = 27$.

31. What is the limit of $f(x) = \frac{2x - 6}{x - 3}$ as $x$ approaches 3?

    2. Factor: $\frac{2(x-3)}{x-3} = 2$, then substitute $x = 3$.

32. What is the limit of $f(x) = \frac{x^2 - 1}{x^2 + 1}$ as $x$ approaches infinity?

    1. Divide by highest power: $\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} \to 1$.

33. Evaluate the limit of $\frac{x^2 - 25}{x - 5}$ as $x$ approaches 5.

    10. Factor: $\frac{(x-5)(x+5)}{x-5} = x+5$, then substitute $x = 5$.

34. What is the limit of $f(x) = \frac{x^2 - 9}{x - 3}$ as $x$ approaches 3?

    6. Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, then substitute $x = 3$.

35. Evaluate the limit of $\frac{x^2 - 4x + 4}{x - 2}$ as $x$ approaches 2.

    0. Factor: $\frac{(x-2)^2}{x-2} = x-2$, then substitute $x = 2$.

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