Selecting Procedures for Determining Limits - AP Calculus AB

  1. What is the limit of x2−1x−1\frac{x^2 - 1}{x - 1} as xx approaches 1?
    2. Factor numerator: (x+1)(x−1)x−1=x+1\frac{(x+1)(x-1)}{x-1} = x+1, so limit is 1+1=21+1=2.
  2. What is the limit of ln(x)\text{ln}(x) as xx approaches Infinity?
    Infinity. Logarithm grows without bound as argument increases.
  3. Evaluate the limit: lim⁡x→∞2x+35x−4\lim_{x \to \infty} \frac{2x + 3}{5x - 4}.
    25\frac{2}{5}. Divide coefficients of highest powers: 25\frac{2}{5}.
  4. What is the limit of ex−1\text{e}^x - 1 as xx approaches 0?
    0. Direct substitution: e0−1=1−1=0e^0 - 1 = 1 - 1 = 0.
  5. State the definition of continuity at a point.
    A function ff is continuous at x=cx = c if limx→cf(x)=f(c)\text{lim}_{x \to c} f(x) = f(c). Function is continuous when limit equals function value.
  6. What is the limit of f(x)=1xf(x) = \frac{1}{x} as xx approaches Infinity?
    0. As denominator grows without bound, fraction approaches 0.
  7. State the result of evaluating limx→Infinityx2+3x+25x2−4\text{lim}_{x \to \text{Infinity}} \frac{x^2 + 3x + 2}{5x^2 - 4}.
    15\frac{1}{5}. Divide by highest power: coefficients of x2x^2 give 15\frac{1}{5}.
  8. State the condition for a function to be continuous at x=cx = c.
    lim⁡x→cf(x)=f(c)\lim_{x \to c} f(x) = f(c). Limit must equal function value for continuity.
  9. What is the limit of f(x)=2x3−x+5f(x) = 2x^3 - x + 5 as xx approaches 0?
    5. Direct substitution: 2(0)3−0+5=52(0)^3 - 0 + 5 = 5.
  10. Which rule applies when directly substituting x=cx = c in rational functions?
    Direct Substitution Rule. When function is continuous at cc, simply substitute x=cx=c.
  11. What is the limit of 1/x1/x as xx approaches 0 from the right?
    Infinity. As x→0+x \to 0^+, denominator approaches 0 positively.
  12. What is the limit of sin(x)\text{sin}(x) as xx approaches 0?
    0. Direct substitution: sin⁡(0)=0\sin(0) = 0.
  13. What is the limit of f(x)=3x+5f(x) = 3x + 5 as xx approaches 2?
    11. Direct substitution: 3(2)+5=113(2) + 5 = 11.
  14. Evaluate the limit: limx→3x2−9x−3\text{lim}_{x \to 3} \frac{x^2 - 9}{x - 3}.
    6. Factor: (x+3)(x−3)x−3=x+3\frac{(x+3)(x-3)}{x-3} = x+3, so limit is 3+3=63+3=6.
  15. What is the squeeze theorem used for?
    Determining limits of functions trapped between two other functions. If g(x)≤f(x)≤h(x)g(x) \leq f(x) \leq h(x) and lim⁡g(x)=lim⁡h(x)=L\lim g(x) = \lim h(x) = L, then lim⁡f(x)=L\lim f(x) = L.
  16. What is the limit of f(x)=1xf(x) = \frac{1}{x} as xx approaches ∞\infty?
    0. As denominator grows without bound, fraction approaches 0.
  17. What is the limit of 1x\frac{1}{x} as xx approaches 0 from the left?
    -Infinity. As x→0−x \to 0^-, denominator approaches 0 negatively.
  18. What is the limit of ln(x)\text{ln}(x) as xx approaches 0 from the right?
    -Infinity. Logarithm approaches −∞-\infty as argument approaches 0.
  19. Find the limit: limx→−Infinity(5x2+3x−2)\text{lim}_{x \to -\text{Infinity}} (5x^2 + 3x - 2).
    Infinity. Highest power dominates: 5x25x^2 term grows without bound.
  20. What is the limit of cos(x)\text{cos}(x) as xx approaches pi2\frac{\text{pi}}{2}?
    0. Direct substitution: cos⁡(π2)=0\cos(\frac{\pi}{2}) = 0.
  21. State the condition for using L'Hôpital's Rule.
    Indeterminate forms 00\frac{0}{0} or InfinityInfinity\frac{\text{Infinity}}{\text{Infinity}}. Rule applies only to these specific indeterminate forms.
  22. Evaluate the limit: limx→Infinity(e−x)\text{lim}_{x \to \text{Infinity}} (\text{e}^{-x}).
    0. Exponential decay: e−x→0e^{-x} \to 0 as x→∞x \to \infty.
  23. Which method is used for limits of rational functions with complex roots?
    Factor and Cancel. Factor numerator and denominator, then cancel common factors.
  24. What is the limit of ln(x)x\frac{\text{ln}(x)}{x} as xx approaches Infinity?
    0. L'Hôpital's Rule: lim⁡x→∞ln⁡(x)x=lim⁡x→∞1/x1=0\lim_{x \to \infty} \frac{\ln(x)}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0.
  25. Determine the limit: limx→12x−2x2−1\text{lim}_{x \to 1} \frac{2x - 2}{x^2 - 1}.
    1. Factor denominator: 2(x−1)(x+1)(x−1)=2x+1\frac{2(x-1)}{(x+1)(x-1)} = \frac{2}{x+1}, so limit is 22=1\frac{2}{2}=1.
  26. What is the limit of x2x^2 as xx approaches -3?
    9. Direct substitution: (−3)2=9(-3)^2 = 9.
  27. Determine the limit: limx→2x2−4x−2\text{lim}_{x \to 2} \frac{x^2 - 4}{x - 2}.
    4. Factor: (x+2)(x−2)x−2=x+2\frac{(x+2)(x-2)}{x-2} = x+2, so limit is 2+2=42+2=4.
  28. What is the limit of tan(x)\text{tan}(x) as xx approaches pi4\frac{\text{pi}}{4}?
    1. Direct substitution: tan⁡(π4)=1\tan(\frac{\pi}{4}) = 1.
  29. Evaluate the limit: limx→0sin(2x)x\text{lim}_{x \to 0} \frac{\text{sin}(2x)}{x}.
    2. Use lim⁡x→0sin⁡(x)x=1\lim_{x \to 0} \frac{\sin(x)}{x} = 1: sin⁡(2x)x=2⋅sin⁡(2x)2x=2(1)=2\frac{\sin(2x)}{x} = 2 \cdot \frac{\sin(2x)}{2x} = 2(1) = 2.
  30. Find the limit: limx→Infinity(3−5x)\text{lim}_{x \to \text{Infinity}} (3 - \frac{5}{x}).
    3. As x→∞x \to \infty, 5x→0\frac{5}{x} \to 0, so limit is 3−0=33-0=3.
  31. Which theorem states that if ff is continuous on [a,b][a, b], then ff takes every value between f(a)f(a) and f(b)f(b)?
    Intermediate Value Theorem. Guarantees continuous functions achieve all intermediate values.
  32. What is the limit of exe^x as xx approaches 0?
    1. Direct substitution: e0=1e^0 = 1.
  33. What is the limit of f(x)=x2−1x2+1f(x) = \frac{x^2 - 1}{x^2 + 1} as xx approaches Infinity?
    1. Divide by highest power: x2x2=1\frac{x^2}{x^2} = 1 as x→∞x \to \infty.
  34. Identify the procedure for limits with polynomials.
    Use Direct Substitution if possible. Polynomials are continuous everywhere, so substitute directly.
  35. State the L'Hôpital's Rule.
    If lim⁡x→cf(x)g(x)=00\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0}, then lim⁡x→cf(x)g(x)=lim⁡x→cf′(x)g′(x)\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}. Apply when limit gives 00\frac{0}{0} or ∞∞\frac{\infty}{\infty} form.
  36. Find the limit: limx→1(3x2+2x−1)\text{lim}_{x \to 1} (3x^2 + 2x - 1).
    4. Direct substitution: 3(1)2+2(1)−1=43(1)^2 + 2(1) - 1 = 4.
  37. Evaluate the limit: limx→01−cos(x)x2\text{lim}_{x \to 0} \frac{1 - \text{cos}(x)}{x^2}.
    12\frac{1}{2}. Standard limit: lim⁡x→01−cos⁡(x)x2=12\lim_{x \to 0} \frac{1-\cos(x)}{x^2} = \frac{1}{2}.

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