AP Calculus AB Quiz: Algebraic Properties Of Limits
20 questions · exam conditions
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Algebraic Properties Of LimitsQuestion 1 of 20

Let f and g be functions such that lim⁡x→−2f(x)g(x)=−3\lim_{x \to -2} \frac{f(x)}{g(x)} = -3 and lim⁡x→−2f(x)=6\lim_{x \to -2} f(x) = 6. What must be the value of lim⁡x→−2g(x)\lim_{x \to -2} g(x)?

−18-18
−2-2
22
The limit cannot be determined.
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AP Calculus AB Quiz

AP Calculus AB Quiz: Algebraic Properties Of Limits

Practice Algebraic Properties Of Limits in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Algebraic Properties Of Limits, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let f and g be functions such that lim⁡x→−2f(x)g(x)=−3\lim_{x \to -2} \frac{f(x)}{g(x)} = -3 and lim⁡x→−2f(x)=6\lim_{x \to -2} f(x) = 6. What must be the value of lim⁡x→−2g(x)\lim_{x \to -2} g(x)?

  1. −18-18
  2. −2-2 (correct answer)
  3. 22
  4. The limit cannot be determined.
Explanation: Let L=lim⁡x→−2g(x)L = \lim_{x \to -2} g(x). The quotient property states lim⁡x→−2f(x)lim⁡x→−2g(x)=−3\frac{\lim_{x \to -2} f(x)}{\lim_{x \to -2} g(x)} = -3. Substituting the known value gives 6L=−3\frac{6}{L} = -3. Solving for L yields L=6−3=−2L = \frac{6}{-3} = -2. Distractor A is the result of multiplication instead of division. Distractor C is a sign error.

Question 2

Let f(x)=2x+3f(x)=2^x+3. Evaluate lim⁡x→2f(x)\lim_{x\to 2} f(x) using limit laws.

  1. 8
  2. 7 (correct answer)
  3. 5
  4. 6
  5. 9
Explanation: This exponential function is continuous everywhere, allowing direct substitution. Using limit laws: lim⁡x→2f(x)=22+3=4+3=7\lim_{x\to 2} f(x) = 2^2 + 3 = 4 + 3 = 7. We apply the sum rule, with the exponential term 22=42^2 = 4 and the constant term 3. A common mistake would be confusing exponential and polynomial notation or making arithmetic errors. The transferable strategy is to recognize that exponential functions are continuous everywhere, so direct substitution applies, and to compute exponential values carefully.

Question 3

If f(x)=12x2+4f(x)=\frac{1}{2}x^2+4, evaluate lim⁡x→2f(x)\lim_{x\to 2} f(x) using limit laws.

  1. 6
  2. 8 (correct answer)
  3. 4
  4. 2
  5. 12
Explanation: This quadratic function with a fractional coefficient is continuous at x=2x = 2. Using direct substitution: lim⁡x→2f(x)=12(2)2+4=12(4)+4=2+4=8\lim_{x\to 2} f(x) = \frac{1}{2}(2)^2 + 4 = \frac{1}{2}(4) + 4 = 2 + 4 = 8. We apply the sum rule and scalar multiplication rule, carefully computing 12⋅4=2\frac{1}{2} \cdot 4 = 2. A common mistake would be incorrectly handling the fractional coefficient or making arithmetic errors. The transferable strategy is to work step-by-step with fractional coefficients, ensuring proper order of operations in polynomial functions.

Question 4

Let f and g be functions such that lim⁡x→2g(x)=3\lim_{x \to 2} g(x) = 3. If f is continuous at x=3x=3 and f(3)=5f(3) = 5, what is lim⁡x→2f(g(x))\lim_{x \to 2} f(g(x))?

  1. 22
  2. 33
  3. 55 (correct answer)
  4. 88
Explanation: Because f is continuous at lim⁡x→2g(x)=3\lim_{x \to 2} g(x) = 3, we can use the property for the limit of a composite function: lim⁡x→2f(g(x))=f(lim⁡x→2g(x))=f(3)=5\lim_{x \to 2} f(g(x)) = f(\lim_{x \to 2} g(x)) = f(3) = 5. Distractor B is the limit of the inner function, not the composite function. Distractor D is the sum of the input and output values (3+53+5).

Question 5

If lim⁡x→−2f(x)=3\lim_{x \to -2} f(x) = 3 and lim⁡x→−2g(x)=−1\lim_{x \to -2} g(x) = -1, what is lim⁡x→−2(2[f(x)]2g(x))\lim_{x \to -2} (2[f(x)]^2 g(x))?

  1. −18-18 (correct answer)
  2. −12-12
  3. 1818
  4. −6-6
Explanation: Using the constant multiple, power, and product properties of limits: lim⁡x→−2(2[f(x)]2g(x))=2⋅(lim⁡x→−2f(x))2⋅(lim⁡x→−2g(x))=2⋅(3)2⋅(−1)=2⋅9⋅(−1)=−18\lim_{x \to -2} (2[f(x)]^2 g(x)) = 2 \cdot (\lim_{x \to -2} f(x))^2 \cdot (\lim_{x \to -2} g(x)) = 2 \cdot (3)^2 \cdot (-1) = 2 \cdot 9 \cdot (-1) = -18. Distractor D results from ignoring the square on f(x) (2⋅3⋅(−1)=−62 \cdot 3 \cdot (-1) = -6). Distractor C is a sign error. Distractor B may result from incorrectly applying the power (2⋅(2⋅3)⋅(−1)=−122 \cdot (2 \cdot 3) \cdot (-1) = -12).

Question 6

Suppose f and g are functions such that lim⁡x→1f(x)g(x)=3\lim_{x \to 1} \frac{f(x)}{g(x)} = 3 and lim⁡x→1(f(x)+g(x))=8\lim_{x \to 1} (f(x) + g(x)) = 8. What is the value of lim⁡x→1f(x)\lim_{x \to 1} f(x)?

  1. 22
  2. 33
  3. 66 (correct answer)
  4. 1212
Explanation: Let L=lim⁡x→1f(x)L = \lim_{x \to 1} f(x) and M=lim⁡x→1g(x)M = \lim_{x \to 1} g(x). The given information creates a system of equations: LM=3\frac{L}{M} = 3 and L+M=8L+M=8. From the first equation, L=3ML = 3M. Substituting into the second gives 3M+M=83M + M = 8, which means 4M=84M=8 and M=2M=2. Therefore, L=3M=3(2)=6L = 3M = 3(2) = 6. Distractor A is the value for the limit of g(x). Distractor B is the value of the quotient. Distractor D would be the answer if the second equation was f(x)−g(x)=8f(x)-g(x)=8.

Question 7

Given p(x)=6−x3p(x)=\frac{6-x}{3}, find lim⁡x→0p(x)\lim_{x\to 0} p(x) using limit laws.

  1. 0
  2. 3
  3. 6
  4. 2 (correct answer)
  5. -2
Explanation: This rational function with a constant denominator is continuous at x=0x = 0. Using direct substitution: lim⁡x→0p(x)=6−03=63=2\lim_{x\to 0} p(x) = \frac{6 - 0}{3} = \frac{6}{3} = 2. The function is linear in the numerator with a constant denominator, making it continuous everywhere. A common error would be arithmetic mistakes in 6−0=66 - 0 = 6 or 63=2\frac{6}{3} = 2. The key strategy is to recognize that rational functions with constant non-zero denominators are continuous everywhere, allowing direct substitution.

Question 8

Given g(x)=x5+25g(x)=\frac{x}{5}+\frac{2}{5}, evaluate lim⁡x→8g(x)\lim_{x\to 8} g(x) using limit laws.

  1. 85\frac{8}{5}
  2. 2 (correct answer)
  3. 125\frac{12}{5}
  4. 25\frac{2}{5}
  5. 105\frac{10}{5}
Explanation: This linear function with fractional coefficients is continuous everywhere. Using direct substitution: lim⁡x→8g(x)=85+25=8+25=105=2\lim_{x\to 8} g(x) = \frac{8}{5} + \frac{2}{5} = \frac{8 + 2}{5} = \frac{10}{5} = 2. We can also factor as g(x)=x+25g(x) = \frac{x + 2}{5}, giving 8+25=105=2\frac{8 + 2}{5} = \frac{10}{5} = 2. A common mistake would be not combining fractions properly or arithmetic errors. The transferable approach is to recognize when expressions can be simplified by factoring or by combining like terms with common denominators.

Question 9

A function is M(x)=25+4xM(x)=\sqrt{25+4x}. Evaluate lim⁡x→0M(x)\lim_{x\to 0} M(x) using limit laws.

  1. 5 (correct answer)
  2. 4\sqrt{4}
  3. 25
  4. 10
  5. 29\sqrt{29}
Explanation: This square root function is continuous at x=0x = 0 since 25+4(0)=25>025 + 4(0) = 25 > 0. Using direct substitution: lim⁡x→0M(x)=25+4(0)=25+0=25=5\lim_{x\to 0} M(x) = \sqrt{25 + 4(0)} = \sqrt{25 + 0} = \sqrt{25} = 5. The expression under the radical is positive, ensuring the function is well-defined. A common mistake would be arithmetic errors in 4(0)=04(0) = 0 or 25=5\sqrt{25} = 5. The key strategy is to verify that expressions under radicals remain non-negative, then apply direct substitution with careful arithmetic.

Question 10

A population model is N(t)=100(1.02)tN(t)=100(1.02)^t. Find lim⁡t→0N(t)\lim_{t\to 0} N(t) using limit laws.

  1. 102
  2. 2
  3. 100 (correct answer)
  4. 98
  5. 1.02
Explanation: This exponential function with a coefficient is continuous everywhere. Using direct substitution: lim⁡t→0N(t)=100(1.02)0=100(1)=100\lim_{t\to 0} N(t) = 100(1.02)^0 = 100(1) = 100. We apply the property that any non-zero number raised to the power 0 equals 1, specifically (1.02)0=1(1.02)^0 = 1. A common error would be forgetting that a0=1a^0 = 1 for any a≠0a \neq 0 or making mistakes with the coefficient. The key strategy is to remember fundamental exponent rules and recognize that exponential functions are continuous, allowing direct substitution.

Question 11

For g(x)=ln⁡(x)+1g(x)=\ln(x)+1, find lim⁡x→1g(x)\lim_{x\to 1} g(x) using limit laws.

  1. 0
  2. 2
  3. 1 (correct answer)
  4. ln⁡(2)\ln(2)
  5. ln⁡(1)\ln(1)
Explanation: This logarithmic function is continuous at x=1x = 1 since the argument is positive. Using direct substitution: lim⁡x→1g(x)=ln⁡(1)+1=0+1=1\lim_{x\to 1} g(x) = \ln(1) + 1 = 0 + 1 = 1. We apply the sum rule, noting that ln⁡(1)=0\ln(1) = 0 by the definition of natural logarithm. A common error would be incorrectly recalling that ln⁡(1)=1\ln(1) = 1 instead of ln⁡(1)=0\ln(1) = 0, or not checking the domain of the logarithm. The key strategy is to memorize fundamental logarithmic values and verify that the argument is positive before applying direct substitution.

Question 12

If f(x)=23x+13f(x)=\frac{2}{3}x+\frac{1}{3}, evaluate lim⁡x→3f(x)\lim_{x\to 3} f(x) using limit laws.

  1. 73\frac{7}{3} (correct answer)
  2. 2
  3. 53\frac{5}{3}
  4. 13\frac{1}{3}
  5. 103\frac{10}{3}
Explanation: This linear function with fractional coefficients requires careful arithmetic. Using direct substitution: lim⁡x→3f(x)=23(3)+13=63+13=2+13=63+13=73\lim_{x\to 3} f(x) = \frac{2}{3}(3) + \frac{1}{3} = \frac{6}{3} + \frac{1}{3} = 2 + \frac{1}{3} = \frac{6}{3} + \frac{1}{3} = \frac{7}{3}. We apply the sum rule and scalar multiplication rule, being careful with fraction arithmetic. A common mistake would be incorrectly computing 23⋅3=2\frac{2}{3} \cdot 3 = 2 or making errors when adding fractions. The transferable approach is to work systematically with fractional coefficients, ensuring proper arithmetic at each step.

Question 13

A temperature model is T(t)=t3−2tT(t)=t^3-2t. Evaluate lim⁡t→−1T(t)\lim_{t\to -1} T(t).

  1. 3
  2. 1 (correct answer)
  3. -3
  4. -1
  5. 0
Explanation: This cubic function is continuous everywhere, allowing direct substitution. Using limit laws: lim⁡t→−1T(t)=(−1)3−2(−1)=−1+2=1\lim_{t\to -1} T(t) = (-1)^3 - 2(-1) = -1 + 2 = 1. We apply the difference rule and power rule, carefully handling the negative input: (−1)3=−1(-1)^3 = -1 and −2(−1)=+2-2(-1) = +2. A common mistake would be sign errors when working with negative values, particularly (−1)3=−1(-1)^3 = -1 or the double negative in −2(−1)-2(-1). The transferable strategy is to work methodically with signs when substituting negative values into polynomial expressions.

Question 14

A function is w(x)=3xx+1w(x)=\frac{3x}{x+1}. Evaluate lim⁡x→2w(x)\lim_{x\to 2} w(x) using limit laws.

  1. 2 (correct answer)
  2. 3
  3. 32\frac{3}{2}
  4. 53\frac{5}{3}
  5. 13\frac{1}{3}
Explanation: This rational function is continuous at x=2x = 2 since 2+1=3≠02 + 1 = 3 \neq 0. Using the quotient rule: lim⁡x→2w(x)=lim⁡x→2(3x)lim⁡x→2(x+1)=3(2)2+1=63=2\lim_{x\to 2} w(x) = \frac{\lim_{x\to 2}(3x)}{\lim_{x\to 2}(x + 1)} = \frac{3(2)}{2 + 1} = \frac{6}{3} = 2. Both numerator and denominator limits exist, and the denominator is non-zero, allowing direct application of the quotient rule. A common mistake would be not checking that the denominator is non-zero or making arithmetic errors in the fraction simplification. The transferable approach is to verify continuity, then apply the quotient rule systematically.

Question 15

A function is r(x)=7+x3r(x)=7+\frac{x}{3}. Find lim⁡x→6r(x)\lim_{x\to 6} r(x) using limit laws.

  1. 9 (correct answer)
  2. 7
  3. 13
  4. 2
  5. 8
Explanation: This linear function with a fractional coefficient is continuous everywhere. Using direct substitution: lim⁡x→6r(x)=7+63=7+2=9\lim_{x\to 6} r(x) = 7 + \frac{6}{3} = 7 + 2 = 9. We apply the sum rule for limits, evaluating each term separately: the constant term gives 7, and the rational term gives 63=2\frac{6}{3} = 2. A common error would be arithmetic mistakes in the fraction division or not recognizing this as a continuous function. The key strategy is to identify continuous functions and apply direct substitution, treating each term according to the appropriate limit law.

Question 16

If F(x)=x2−45+2F(x)=\frac{x^2-4}{5}+2, evaluate lim⁡x→1F(x)\lim_{x\to 1} F(x) using limit laws.

  1. 15\frac{1}{5}
  2. 75\frac{7}{5} (correct answer)
  3. 115\frac{11}{5}
  4. 65\frac{6}{5}
  5. 95\frac{9}{5}
Explanation: This rational function with a quadratic numerator requires the quotient rule. Since the denominator is non-zero at x=1x = 1: lim⁡x→1F(x)=(1)2−45+2=1−45+2=−35+2=−35+105=75\lim_{x\to 1} F(x) = \frac{(1)^2 - 4}{5} + 2 = \frac{1 - 4}{5} + 2 = \frac{-3}{5} + 2 = -\frac{3}{5} + \frac{10}{5} = \frac{7}{5}. We apply the sum rule, evaluating the rational term and constant term separately. A common mistake would be sign errors when computing 1−4=−31 - 4 = -3 or incorrectly adding fractions. The transferable approach is to handle each term according to the appropriate limit law and work carefully with negative values and fraction arithmetic.

Question 17

If lim⁡x→−1f(x)=4\lim_{x \to -1} f(x) = 4 and lim⁡x→−1g(x)=−3\lim_{x \to -1} g(x) = -3, what is lim⁡x→−1[f(x)⋅g(x)]\lim_{x \to -1} [f(x) \cdot g(x)]?

  1. 11
  2. −12-12 (correct answer)
  3. 1212
  4. 77
Explanation: By the product property of limits, lim⁡x→−1[f(x)⋅g(x)]=(lim⁡x→−1f(x))⋅(lim⁡x→−1g(x))=(4)⋅(−3)=−12\lim_{x \to -1} [f(x) \cdot g(x)] = (\lim_{x \to -1} f(x)) \cdot (\lim_{x \to -1} g(x)) = (4) \cdot (-3) = -12. Distractor A incorrectly uses the sum property (4+(−3)=14 + (-3) = 1). Distractor C ignores the negative sign on the limit of g(x) (4⋅3=124 \cdot 3 = 12). Distractor D incorrectly uses the difference property (4−(−3)=74 - (-3) = 7).

Question 18

Let f and g be functions such that lim⁡x→af(x)=L\lim_{x \to a} f(x) = L and lim⁡x→ag(x)=M\lim_{x \to a} g(x) = M. The property lim⁡x→af(x)g(x)=LM\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M} holds true under which of the following necessary conditions?

  1. MM can be any real number.
  2. L=0L = 0.
  3. M≠0M \neq 0. (correct answer)
  4. L≠0L \neq 0 and M≠0M \neq 0.
Explanation: The quotient property for limits is only guaranteed to hold if the limit of the denominator is not zero. Therefore, the required condition is M=lim⁡x→ag(x)≠0M = \lim_{x \to a} g(x) \neq 0. If M=0M=0, the limit of the quotient may be infinite or may not exist. The value of L does not affect the validity of the property itself.

Question 19

Let f and g be functions such that lim⁡x→3f(x)=5\lim_{x \to 3} f(x) = 5 and lim⁡x→3g(x)=−2\lim_{x \to 3} g(x) = -2. What is the value of lim⁡x→3[2f(x)−g(x)]\lim_{x \to 3} [2f(x) - g(x)]?

  1. 33
  2. 77
  3. 88
  4. 1212 (correct answer)
Explanation: Using the sum, difference, and constant multiple properties of limits: lim⁡x→3[2f(x)−g(x)]=2lim⁡x→3f(x)−lim⁡x→3g(x)=2(5)−(−2)=10+2=12\lim_{x \to 3} [2f(x) - g(x)] = 2 \lim_{x \to 3} f(x) - \lim_{x \to 3} g(x) = 2(5) - (-2) = 10 + 2 = 12.

Question 20

Given that lim⁡x→5f(x)=10\lim_{x \to 5} f(x) = 10 and lim⁡x→5g(x)=−2\lim_{x \to 5} g(x) = -2, find lim⁡x→5f(x)g(x)\lim_{x \to 5} \frac{f(x)}{g(x)}.

  1. −5-5 (correct answer)
  2. −20-20
  3. 88
  4. The limit does not exist.
Explanation: Using the quotient property of limits, since lim⁡x→5g(x)≠0\lim_{x \to 5} g(x) \neq 0, we have lim⁡x→5f(x)g(x)=lim⁡x→5f(x)lim⁡x→5g(x)=10−2=−5\lim_{x \to 5} \frac{f(x)}{g(x)} = \frac{\lim_{x \to 5} f(x)}{\lim_{x \to 5} g(x)} = \frac{10}{-2} = -5. Distractor B is the result of applying the product property. Distractor C is the result of applying the sum property. Distractor D is incorrect because the limit of the denominator is not zero.