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

A concentration is k(x)=x2−4x+1xk(x)=\dfrac{x^2-4x+1}{x}; find lim⁡x→5k(x)\lim_{x\to 5} k(x).

25−20+15\dfrac{25-20+1}{5}
(5−4)2+15\dfrac{(5-4)^2+1}{5}
25−20+1525-20+\dfrac{1}{5}
25−4(5)+10\dfrac{25-4(5)+1}{0}
52−4+15\dfrac{5^2-4+1}{5}
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AP Calculus BC Quiz

AP Calculus BC Quiz: Algebraic Properties Of Limits

Practice Algebraic Properties Of Limits in AP Calculus BC 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 BC.

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

A concentration is k(x)=x2−4x+1xk(x)=\dfrac{x^2-4x+1}{x}; find lim⁡x→5k(x)\lim_{x\to 5} k(x).

  1. 25−20+15\dfrac{25-20+1}{5} (correct answer)
  2. (5−4)2+15\dfrac{(5-4)^2+1}{5}
  3. 25−20+1525-20+\dfrac{1}{5}
  4. 25−4(5)+10\dfrac{25-4(5)+1}{0}
  5. 52−4+15\dfrac{5^2-4+1}{5}
Explanation: This problem requires using the limit laws for rational functions. The denominator at x=5 is 5 ≠0, so direct substitution is applicable. The limit is (525^2 -4*5 +1)/5 = (25-20+1)/5 =6/5. The choice A is this expression. A tempting distractor is D dfrac{25-4(5)+1}{0}, perhaps from incorrectly using 0 in denominator. A transferable strategy is to substitute the value into both numerator and denominator separately and then divide.

Question 2

As tt approaches 22 seconds, the sensor output is S(t)=3t2−5t+4S(t)=3t^2-5t+4; find lim⁡t→2S(t)\lim_{t\to 2} S(t).

  1. 66
  2. 22
  3. 6(2)2−5(2)+46(2)^2-5(2)+4
  4. 3(2)2−5(2)+43(2)^2-5(2)+4 (correct answer)
  5. 3(2)2−5+43(2)^2-5+4
Explanation: This problem requires using the limit laws, specifically the direct substitution property for polynomials. Since S(t) is a polynomial, it is continuous, and the limit as t approaches 2 is S(2) = 3(2)^2 -5(2) +4. This expression simplifies to 12 -10 +4 =6. The choice D is this exact expression for direct substitution. A tempting distractor is C, which uses 6 as the coefficient instead of 3, perhaps from misreading 3t^2 as 6t^2, leading to an incorrect calculation. A transferable strategy is to identify if the function is continuous at the point and substitute directly for the limit.

Question 3

A signal is modeled by f(x)=5−2xx2+1f(x)=\dfrac{5-2x}{x^2+1}. Find lim⁡x→0f(x)\lim_{x\to 0} f(x).

  1. 55 (correct answer)
  2. 00
  3. 52\dfrac{5}{2}
  4. −5-5
  5. 31\dfrac{3}{1}
Explanation: This problem requires using limit laws on a rational function where the denominator is non-zero at the limit point. Since 02+1=1≠00^2+1 = 1 \neq 0, we can apply direct substitution: lim⁡x→05−2xx2+1=5−2(0)02+1=5−00+1=51=5\lim_{x\to 0} \frac{5-2x}{x^2+1} = \frac{5-2(0)}{0^2+1} = \frac{5-0}{0+1} = \frac{5}{1} = 5. The limit laws allow us to evaluate the numerator and denominator separately when the denominator limit is non-zero. A potential error would be to think the limit is 0 because xx approaches 0, forgetting that we need to evaluate the entire expression. For rational functions, always substitute the limit value into both numerator and denominator, checking that the denominator remains non-zero.

Question 4

Given R(x)=(x+1)(x−3)x−3R(x)=\dfrac{(x+1)(x-3)}{x-3} for x≠3x\ne 3, evaluate lim⁡x→3R(x)\lim_{x\to 3} R(x).

  1. (3+1)(3−3)3−3\dfrac{(3+1)(3-3)}{3-3}
  2. 3+13+1 (correct answer)
  3. (3+1)(3−3)3\dfrac{(3+1)(3-3)}{3}
  4. (3+1)(3)3−3\dfrac{(3+1)(3)}{3-3}
  5. (3)(3−3)3−3\dfrac{(3)(3-3)}{3-3}
Explanation: This problem requires using the algebraic properties of limits, including simplifying rational expressions before taking the limit. The function R(x) = (x+1)(x-3)/(x-3) simplifies to x+1 for x≠3. Therefore, the limit as x approaches 3 is 3+1=4. The choice B is this value. A tempting distractor is A, which plugs into the original to get 0/0, an indeterminate form that doesn't give the limit. A transferable strategy is to factor and cancel common factors in rational functions when direct substitution gives 0/0.

Question 5

A growth function is P(x)=7−3xP(x)=7-\dfrac{3}{x}. Determine lim⁡x→6P(x)\lim_{x\to 6} P(x).

  1. 132\dfrac{13}{2} (correct answer)
  2. 112\dfrac{11}{2}
  3. 72\dfrac{7}{2}
  4. 43\dfrac{4}{3}
  5. 37\dfrac{3}{7}
Explanation: This problem uses limit laws to evaluate a function involving subtraction of a rational term. Since x=6x=6 makes the denominator non-zero, we can use direct substitution: lim⁡x→6(7−3x)=7−36=7−12=142−12=132\lim_{x\to 6} \left(7-\frac{3}{x}\right) = 7-\frac{3}{6} = 7-\frac{1}{2} = \frac{14}{2}-\frac{1}{2} = \frac{13}{2}. The limit laws tell us we can evaluate the limit of a difference by taking the difference of the individual limits. A potential mistake would be to compute 7−3=47-3 = 4 and then divide by 6, getting 46=23\frac{4}{6} = \frac{2}{3}, by misapplying the order of operations. For expressions with multiple operations, always follow the correct order: division before subtraction.

Question 6

A particle's position is s(t)=t2+1t+1s(t)=\dfrac{t^2+1}{t+1}. Determine lim⁡t→3s(t)\lim_{t\to 3} s(t).

  1. 52\dfrac{5}{2} (correct answer)
  2. 103\dfrac{10}{3}
  3. 104\dfrac{10}{4}
  4. 45\dfrac{4}{5}
  5. 34\dfrac{3}{4}
Explanation: This problem uses limit laws to evaluate a rational function where direct substitution is valid. Since the denominator t+1t+1 is non-zero when t=3t=3, we can substitute directly: lim⁡t→3t2+1t+1=32+13+1=9+14=104=52\lim_{t\to 3} \frac{t^2+1}{t+1} = \frac{3^2+1}{3+1} = \frac{9+1}{4} = \frac{10}{4} = \frac{5}{2}. The limit laws allow us to evaluate the numerator and denominator limits separately when the denominator limit is non-zero. A common error would be to simplify incorrectly, perhaps getting 103\frac{10}{3} by mistakenly using tt instead of t+1t+1 in the denominator. For rational functions, always verify the denominator is non-zero at the limit point before applying direct substitution.

Question 7

If m(x)=3x−6x+5+2m(x)=\dfrac{3x-6}{x+5}+2, find lim⁡x→1m(x)\lim_{x\to 1} m(x).

  1. 3(1)−61+5+2\dfrac{3(1)-6}{1+5}+2 (correct answer)
  2. 3(1−6)1+5+2\dfrac{3(1-6)}{1+5}+2
  3. 3−61+5+2\dfrac{3-6}{1}+5+2
  4. 3(1)−61+5+2\dfrac{3(1)-6}{1}+5+2
  5. 3(1)−61+5+2\dfrac{3(1)-6}{1+5+2}
Explanation: This problem requires using the limit laws for sums and quotients. The function is continuous at x=1 since denominator 1+5=6 ≠0. The limit is [3(1)-6]/(1+5) +2 = (-3)/6 +2 = -0.5 +2=1.5. The choice A is this expression. A tempting distractor is B dfrac{3(1-6)}{1+5}+2 = 3*(-5)/6 +2 = -15/6 +2 = -2.5+2=-0.5, from misplacing the parentheses in numerator. A transferable strategy is to apply substitution to each component and combine using the sum law.

Question 8

During a lab, g(x)=2x2+3x−5x−1g(x)=\dfrac{2x^2+3x-5}{x-1} is recorded; evaluate lim⁡x→1(2x2+3x−5)\lim_{x\to 1} (2x^2+3x-5).

  1. 00 (correct answer)
  2. −4-4
  3. 55
  4. −2-2
  5. 44
Explanation: This problem illustrates the use of limit laws, specifically the sum, difference, and constant multiple rules for limits. The expression 2x^2 + 3x - 5 is a polynomial, so the limit as x approaches 1 is simply its value at x=1. Substituting x=1 gives 2*(1)^2 + 3*1 - 5 = 2 + 3 - 5 = 0. This follows from applying the power rule to each term and then using the sum and difference properties of limits. A tempting distractor like -4 might come from confusing it with evaluating g(x), but the question asks only for the numerator, which is defined at x=1. A transferable strategy is to apply direct substitution for polynomials, as they are continuous everywhere.

Question 9

A function is m(x)=52x2−4x+9m(x)=\dfrac{5}{2}x^2-4x+9; evaluate lim⁡x→0m(x)\lim_{x\to 0} m(x).

  1. 99 (correct answer)
  2. 00
  3. −9-9
  4. 52\dfrac{5}{2}
  5. −4-4
Explanation: This problem illustrates the use of limit laws, specifically the sum, difference, and constant multiple rules. m(x) is a polynomial, continuous everywhere, so the limit as x approaches 0 is m(0). Substituting x=0 gives (5/2)0 - 40 + 9 = 9. This follows from applying the power rule to each term and combining with sum properties. A tempting distractor like 0 might come from only considering the variable terms, but the constant 9 remains. A transferable strategy is to evaluate polynomials at the limit point directly due to their continuity.

Question 10

If f(x)=5x−12x+3f(x)=\dfrac{5x-1}{2x+3} models efficiency, determine lim⁡x→−1f(x)\lim_{x\to -1} f(x).

  1. 5(−1)−12(−1)+3\dfrac{5(-1)-1}{2(-1)+3}
  2. 5(−1)−12(−1)+3\dfrac{5(-1)-1}{2(-1)}+3
  3. 5−12+3\dfrac{5-1}{2+3}
  4. −5−1−2+3\dfrac{-5-1}{-2+3} (correct answer)
  5. −6−1\dfrac{-6}{-1}
Explanation: This problem requires using the limit laws for quotients. The denominator at x=-1 is 2*(-1)+3=-2+3=1 ≠0, so direct substitution works. The limit is (5*(-1)-1)/(2*(-1)+3) = (-5-1)/(-2+3) = -6/1 = -6. The choice D represents this expression with the signs correctly applied. A tempting distractor is C dfrac{5-1}{2+3}=4/5, which ignores the negative in x=-1. A transferable strategy is to carefully plug in the value with attention to negative signs and order of operations.

Question 11

For g(h)=(h+2)34h−1g(h)=\dfrac{(h+2)^3}{4h-1} in a lab model, evaluate lim⁡h→1g(h)\lim_{h\to 1} g(h).

  1. 99 (correct answer)
  2. 2727
  3. 274\dfrac{27}{4}
  4. 273\dfrac{27}{3}
  5. 327\dfrac{3}{27}
Explanation: This limit involves applying limit laws to a rational function with a cubic numerator. First, we verify the denominator is non-zero: 4(1)−1=3≠04(1)-1 = 3 \neq 0, so direct substitution is valid. Computing: lim⁡h→1(h+2)34h−1=(1+2)34(1)−1=333=273=9\lim_{h\to 1} \frac{(h+2)^3}{4h-1} = \frac{(1+2)^3}{4(1)-1} = \frac{3^3}{3} = \frac{27}{3} = 9. The limit laws allow us to evaluate powers and quotients when all operations are defined. A common error would be to compute (1+2)3=9(1+2)^3 = 9 instead of 2727, confusing squaring with cubing. When evaluating limits involving powers, carefully compute each exponent and simplify the resulting fraction completely.

Question 12

For a revenue function R(x)=2x2−3x+1xR(x)=\dfrac{2x^2-3x+1}{x}, evaluate lim⁡x→4R(x)\lim_{x\to 4} R(x).

  1. 214\dfrac{21}{4} (correct answer)
  2. 2116\dfrac{21}{16}
  3. 2121
  4. 54\dfrac{5}{4}
  5. 34\dfrac{3}{4}
Explanation: This problem involves using limit laws to evaluate a rational function at a point where it's defined. First, we simplify the expression: R(x)=2x2−3x+1x=2x−3+1xR(x) = \frac{2x^2-3x+1}{x} = 2x-3+\frac{1}{x} for x≠0x \neq 0. Since x=4x=4 is in the domain, we can apply direct substitution: lim⁡x→4R(x)=2(4)−3+14=8−3+14=5+14=214\lim_{x\to 4} R(x) = 2(4)-3+\frac{1}{4} = 8-3+\frac{1}{4} = 5+\frac{1}{4} = \frac{21}{4}. A common mistake would be to evaluate only the numerator at x=4x=4 getting 2121, forgetting to divide by 4. For rational functions, always check that the denominator is non-zero at the limit point before using direct substitution.

Question 13

A projectile's height is h(t)=t2+3tt+3h(t)=\dfrac{t^2+3t}{t+3}; compute lim⁡t→1h(t)\lim_{t\to 1} h(t).

  1. 12+31+3\dfrac{1^2+3}{1+3}
  2. 12+3(1)1+3\dfrac{1^2+3(1)}{1+3} (correct answer)
  3. 12+3(1)1+3\dfrac{1^2+3(1)}{1}+3
  4. 12+3(1)3\dfrac{1^2+3(1)}{3}
  5. 11
Explanation: This problem requires using the limit laws for rational functions. Since the denominator at t=1 is 1+3=4 ≠0, direct substitution is valid. The limit is (121^2 +3*1)/(1+3) = (1+3)/4 =1. The choice B is this plugged-in expression. A tempting distractor is D dfrac{1^2 +3(1)}{3} =4/3, perhaps from misplacing the denominator as 3 alone. A transferable strategy is to evaluate the denominator's limit first to ensure it's not zero before applying direct substitution.

Question 14

A temperature adjustment is A(x)=7−2x+4A(x)=7-\dfrac{2}{x+4}; compute lim⁡x→0A(x)\lim_{x\to 0} A(x).

  1. 7−247-\dfrac{2}{4} (correct answer)
  2. 7−24\dfrac{7-2}{4}
  3. 7−207-\dfrac{2}{0}
  4. 74−2\dfrac{7}{4}-2
  5. 7−2x7-\dfrac{2}{x}
Explanation: This problem requires using the limit laws for constants, quotients, and sums. The function is continuous at x=0, so direct substitution gives 7 - 2/(0+4) =7 -2/4 =7-0.5=6.5. The choice A is this expression. It follows from lim (constant) = constant, and lim 2/(x+4) =2/4. A tempting distractor is C 7 -2/0, perhaps from forgetting the +4 in denominator, leading to undefined. A transferable strategy is to substitute into each part of the expression using the appropriate limit laws.

Question 15

For y(x)=2x+5x2+9y(x)=\dfrac{2x+5}{x^2+9}, compute lim⁡x→−22x+5x2+9\lim_{x\to -2} \dfrac{2x+5}{x^2+9}.

  1. 113\dfrac{1}{13} (correct answer)
  2. −913\dfrac{-9}{13}
  3. 913\dfrac{9}{13}
  4. 15\dfrac{1}{5}
  5. −113\dfrac{-1}{13}
Explanation: This problem illustrates the use of limit laws, including the quotient rule. y(x) = (2x + 5)/(x2x^2 + 9) has denominator at x=-2 of 13 ≠ 0, so direct substitution works. Plugging in x=-2 gives (2*(-2) + 5)/(4 + 9) = 1/13. This uses sum in numerator and denominator before quotient. A tempting distractor like -9/13 might misadd the numerator as -4 -5, but it's +5, yielding positive. A transferable strategy is to carefully compute arithmetic in numerators and denominators for rational limits.

Question 16

A concentration is modeled by c(x)=x2+1x2+4c(x)=\dfrac{x^2+1}{x^2+4}; compute lim⁡x→2c(x)\lim_{x\to 2} c(x).

  1. 58\dfrac{5}{8} (correct answer)
  2. 44\dfrac{4}{4}
  3. 54\dfrac{5}{4}
  4. 12\dfrac{1}{2}
  5. 38\dfrac{3}{8}
Explanation: This problem requires applying the algebraic properties of limits, including the quotient rule. c(x)=x2+1x2+4c(x) = \dfrac{x^2 + 1}{x^2 + 4} has denominator limit at x=2 of 8≠08 \neq 0, so direct substitution applies. Plugging in x=2 gives (4+1)/(4+4)=58(4 + 1)/(4 + 4) = \dfrac{5}{8}. This uses sum rules in numerator and denominator before the quotient. A tempting distractor like 54\dfrac{5}{4} might swap the constants, but correct addition yields 58\dfrac{5}{8}. A transferable strategy is to confirm the denominator's limit is nonzero for rational functions before substituting.

Question 17

A model uses B(x)=1x2+1+x2B(x)=\dfrac{1}{x^2+1}+\dfrac{x}{2}; evaluate lim⁡x→2B(x)\lim_{x\to 2} B(x).

  1. 15+1\dfrac{1}{5}+1 (correct answer)
  2. 14+1\dfrac{1}{4}+1
  3. 15+12\dfrac{1}{5}+\dfrac{1}{2}
  4. 13+1\dfrac{1}{3}+1
  5. 10+1\dfrac{1}{0}+1
Explanation: This problem requires applying the algebraic properties of limits, such as the sum and quotient rules. B(x) = 1/(x2x^2 +1) + x/2 is continuous at x=2, so the limit is B(2). Substituting x=2 gives 1/(4 +1) + 2/2 = 1/5 + 1. This follows from quotient rules for each term and sum property. A tempting distractor like 1/5 + 1/2 might halve incorrectly, but it's 1 + 0.2 = 1.2. A transferable strategy is to compute each addend's limit separately.

Question 18

A temperature model uses T(t)=3t2−5t+8T(t)=3t^2-5t+8. Find lim⁡t→2T(t)\lim_{t\to 2} T(t) using limit laws.

  1. 1010 (correct answer)
  2. 66
  3. −6-6
  4. 00
  5. 1212
Explanation: This problem requires applying limit laws to evaluate the limit of a polynomial function. Since polynomials are continuous everywhere, we can use direct substitution: lim⁡t→2(3t2−5t+8)=3(2)2−5(2)+8=3(4)−10+8=12−10+8=10\lim_{t\to 2} (3t^2-5t+8) = 3(2)^2-5(2)+8 = 3(4)-10+8 = 12-10+8 = 10. The limit laws tell us we can evaluate the limit of each term separately and combine them. A common error would be to substitute t=2t=2 into only part of the expression, perhaps getting 12−10=212-10 = 2 by forgetting the constant term. When evaluating limits of polynomials, always substitute the value directly into the entire expression since polynomials are continuous everywhere.

Question 19

For g(x)=x2+1x2+4g(x)=\dfrac{x^2+1}{x^2+4}, determine lim⁡x→−2g(x)\lim_{x\to -2} g(x) using algebraic limit properties.

  1. (−2)2+1(−2)2+4\dfrac{(-2)^2+1}{(-2)^2+4} (correct answer)
  2. −22+1−22+4\dfrac{-2^2+1}{-2^2+4}
  3. −4+1−4+4\dfrac{-4+1}{-4+4}
  4. 4+14−4\dfrac{4+1}{4-4}
  5. (−2)+1(−2)+4\dfrac{(-2)+1}{(-2)+4}
Explanation: This problem requires using the algebraic limit properties for quotients of polynomials. Since denominator at x=-2 is (-2)^2 +4=4+4=8 ≠0, direct substitution applies. The limit is [(-2)^2 +1]/[(-2)^2 +4] = (4+1)/(4+4)=5/8. The choice A uses the correct order of operations with (-2)^2=4. A tempting distractor is C dfrac{-4+1}{-4+4} = -3/0, from incorrectly computing -2^2 as -4 instead of (+4). A transferable strategy is to remember that (-a)^2 = a^2 is positive, and substitute carefully.

Question 20

A cost adjustment is A(x)=(x−1)(x+5)A(x)=(x-1)(x+5). Compute lim⁡x→−2A(x)\lim_{x\to -2} A(x) using algebraic properties.

  1. −9-9 (correct answer)
  2. 99
  3. −3-3
  4. 33
  5. −7-7
Explanation: This limit problem involves applying limit laws to a factored polynomial expression. Since polynomials are continuous everywhere, we can use direct substitution: lim⁡x→−2(x−1)(x+5)=(−2−1)(−2+5)=(−3)(3)=−9\lim_{x\to -2} (x-1)(x+5) = (-2-1)(-2+5) = (-3)(3) = -9. The limit laws tell us we can evaluate the limit of a product by taking the product of the individual limits. A common mistake would be to get 99 by losing track of the negative signs, computing 3×33 \times 3 instead of (−3)×3(-3) \times 3. When evaluating limits of products, carefully track the sign of each factor, especially when substituting negative values.