AP Calculus AB Quiz: Algebraic Properties Of Limits

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.

Question 1

1. $-18$
2. $-2$ (correct answer)
3. $2$
4. The limit cannot be determined.

Explanation: Let $L = \lim_{x \to -2} g(x)$. The quotient property states $\frac{\lim_{x \to -2} f(x)}{\lim_{x \to -2} g(x)} = -3$. Substituting the known value gives $\frac{6}{L} = -3$. Solving for L yields $L = \frac{6}{-3} = -2$. Distractor A is the result of multiplication instead of division. Distractor C is a sign error.

Question 2

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: $\lim_{x\to 2} f(x) = 2^2 + 3 = 4 + 3 = 7$. We apply the sum rule, with the exponential term $2^2 = 4$ 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

1. 6
2. 8 (correct answer)
3. 4
4. 2
5. 12

Explanation: This quadratic function with a fractional coefficient is continuous at $x = 2$. Using direct substitution: $\lim_{x\to 2} f(x) = \frac{1}{2}(2)^2 + 4 = \frac{1}{2}(4) + 4 = 2 + 4 = 8$. We apply the sum rule and scalar multiplication rule, carefully computing $\frac{1}{2} \cdot 4 = 2$. 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

1. $2$
2. $3$
3. $5$ (correct answer)
4. $8$

Explanation: Because f is continuous at $\lim_{x \to 2} g(x) = 3$, we can use the property for the limit of a composite function: $\lim_{x \to 2} f(g(x)) = f(\lim_{x \to 2} g(x)) = f(3) = 5$. Distractor B is the limit of the inner function, not the composite function. Distractor D is the sum of the input and output values ($3+5$).

Question 5

1. $-18$ (correct answer)
2. $-12$
3. $18$
4. $-6$

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

Question 6

1. $2$
2. $3$
3. $6$ (correct answer)
4. $12$

Explanation: Let $L = \lim_{x \to 1} f(x)$ and $M = \lim_{x \to 1} g(x)$. The given information creates a system of equations: $\frac{L}{M} = 3$ and $L+M=8$. From the first equation, $L = 3M$. Substituting into the second gives $3M + M = 8$, which means $4M=8$ and $M=2$. Therefore, $L = 3M = 3(2) = 6$. 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 $f(x)-g(x)=8$.

Question 7

1. 0
2. 3
3. 6
4. 2 (correct answer)
5. -2

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

Question 8

1. $\frac{8}{5}$
2. 2 (correct answer)
3. $\frac{12}{5}$
4. $\frac{2}{5}$
5. $\frac{10}{5}$

Explanation: This linear function with fractional coefficients is continuous everywhere. Using direct substitution: $\lim_{x\to 8} g(x) = \frac{8}{5} + \frac{2}{5} = \frac{8 + 2}{5} = \frac{10}{5} = 2$. We can also factor as $g(x) = \frac{x + 2}{5}$, giving $\frac{8 + 2}{5} = \frac{10}{5} = 2$. 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

1. 5 (correct answer)
2. $\sqrt{4}$
3. 25
4. 10
5. $\sqrt{29}$

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

Question 10

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: $\lim_{t\to 0} N(t) = 100(1.02)^0 = 100(1) = 100$. We apply the property that any non-zero number raised to the power 0 equals 1, specifically $(1.02)^0 = 1$. A common error would be forgetting that $a^0 = 1$ for any $a \neq 0$ 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

1. 0
2. 2
3. 1 (correct answer)
4. $\ln(2)$
5. $\ln(1)$

Explanation: This logarithmic function is continuous at $x = 1$ since the argument is positive. Using direct substitution: $\lim_{x\to 1} g(x) = \ln(1) + 1 = 0 + 1 = 1$. We apply the sum rule, noting that $\ln(1) = 0$ by the definition of natural logarithm. A common error would be incorrectly recalling that $\ln(1) = 1$ instead of $\ln(1) = 0$, 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

1. $\frac{7}{3}$ (correct answer)
2. 2
3. $\frac{5}{3}$
4. $\frac{1}{3}$
5. $\frac{10}{3}$

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

Question 13

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: $\lim_{t\to -1} T(t) = (-1)^3 - 2(-1) = -1 + 2 = 1$. We apply the difference rule and power rule, carefully handling the negative input: $(-1)^3 = -1$ and $-2(-1) = +2$. A common mistake would be sign errors when working with negative values, particularly $(-1)^3 = -1$ or the double negative in $-2(-1)$. The transferable strategy is to work methodically with signs when substituting negative values into polynomial expressions.

Question 14

1. 2 (correct answer)
2. 3
3. $\frac{3}{2}$
4. $\frac{5}{3}$
5. $\frac{1}{3}$

Explanation: This rational function is continuous at $x = 2$ since $2 + 1 = 3 \neq 0$. Using the quotient rule: $\lim_{x\to 2} w(x) = \frac{\lim_{x\to 2}(3x)}{\lim_{x\to 2}(x + 1)} = \frac{3(2)}{2 + 1} = \frac{6}{3} = 2$. 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

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: $\lim_{x\to 6} r(x) = 7 + \frac{6}{3} = 7 + 2 = 9$. We apply the sum rule for limits, evaluating each term separately: the constant term gives 7, and the rational term gives $\frac{6}{3} = 2$. 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

1. $\frac{1}{5}$
2. $\frac{7}{5}$ (correct answer)
3. $\frac{11}{5}$
4. $\frac{6}{5}$
5. $\frac{9}{5}$

Explanation: This rational function with a quadratic numerator requires the quotient rule. Since the denominator is non-zero at $x = 1$: $\lim_{x\to 1} F(x) = \frac{(1)^2 - 4}{5} + 2 = \frac{1 - 4}{5} + 2 = \frac{-3}{5} + 2 = -\frac{3}{5} + \frac{10}{5} = \frac{7}{5}$. We apply the sum rule, evaluating the rational term and constant term separately. A common mistake would be sign errors when computing $1 - 4 = -3$ 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

1. $1$
2. $-12$ (correct answer)
3. $12$
4. $7$

Explanation: By the product property of limits, $\lim_{x \to -1} [f(x) \cdot g(x)] = (\lim_{x \to -1} f(x)) \cdot (\lim_{x \to -1} g(x)) = (4) \cdot (-3) = -12$. Distractor A incorrectly uses the sum property ($4 + (-3) = 1$). Distractor C ignores the negative sign on the limit of g(x) ($4 \cdot 3 = 12$). Distractor D incorrectly uses the difference property ($4 - (-3) = 7$).

Question 18

1. $M$ can be any real number.
2. $L = 0$.
3. $M \neq 0$. (correct answer)
4. $L \neq 0$ and $M \neq 0$.

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 $M = \lim_{x \to a} g(x) \neq 0$. If $M=0$, 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

1. $3$
2. $7$
3. $8$
4. $12$ (correct answer)

Explanation: Using the sum, difference, and constant multiple properties of limits: $\lim_{x \to 3} [2f(x) - g(x)] = 2 \lim_{x \to 3} f(x) - \lim_{x \to 3} g(x) = 2(5) - (-2) = 10 + 2 = 12$.

Question 20

1. $-5$ (correct answer)
2. $-20$
3. $8$
4. The limit does not exist.

Explanation: Using the quotient property of limits, since $\lim_{x \to 5} g(x) \neq 0$, we have $\lim_{x \to 5} \frac{f(x)}{g(x)} = \frac{\lim_{x \to 5} f(x)}{\lim_{x \to 5} g(x)} = \frac{10}{-2} = -5$. 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.