Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Sign up

Log in

Opening subject page...

Loading your content

Practice

  • All Subjects
  • Algebra Flashcards
  • SAT Math Practice Tests
  • Math Question of the Day
  • Live Classes
  • On-Demand Courses

Varsity Tutors

  • Find a Tutor
  • Test Prep
  • Online Classes
  • K-12 Learning
  • College Search
  • VarsityTutors.com

© 2026 Varsity Tutors. All rights reserved.

← Back to quizzes

AP Calculus AB Quiz

AP Calculus AB Quiz: Determining Limits Using Algebraic Manipulation

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

Question 1 / 20

0 of 20 answered

As xxx approaches 333, a model uses x2−9x−3\frac{x^2-9}{x-3}x−3x2−9​; what is lim⁡x→3x2−9x−3\displaystyle\lim_{x\to 3}\frac{x^2-9}{x-3}x→3lim​x−3x2−9​?

Select an answer to continue

What this quiz covers

This quiz focuses on Determining Limits Using Algebraic Manipulation, 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

As xxx approaches 333, a model uses x2−9x−3\frac{x^2-9}{x-3}x−3x2−9​; what is lim⁡x→3x2−9x−3\displaystyle\lim_{x\to 3}\frac{x^2-9}{x-3}x→3lim​x−3x2−9​?

  1. 000
  2. 666 (correct answer)
  3. 333
  4. 999
  5. Does not exist

Explanation: This limit has the indeterminate form 0/0 when we substitute x = 3 directly, so we need algebraic manipulation. The numerator x² - 9 is a difference of squares that factors as (x - 3)(x + 3). After factoring, we get lim[x→3] [(x - 3)(x + 3)]/(x - 3), and since x approaches but never equals 3, we can cancel the (x - 3) terms. This leaves us with lim[x→3] (x + 3) = 3 + 3 = 6. A common error is thinking the limit doesn't exist because of the 0/0 form, but factoring reveals the removable discontinuity. The strategy is: when you see 0/0, look for common factors to cancel before evaluating.

Question 2

Evaluate lim⁡x→1x2−3x+2x−1\lim\limits_{x\to 1}\dfrac{x^2-3x+2}{x-1}x→1lim​x−1x2−3x+2​, an indeterminate form removable by factoring.

  1. −2-2−2
  2. 222
  3. 000
  4. −1-1−1 (correct answer)
  5. 111

Explanation: This limit initially gives 0/0, requiring factorization of the quadratic numerator. The expression x² - 3x + 2 factors as (x - 1)(x - 2) by finding two numbers that multiply to 2 and add to -3. The limit becomes (x - 1)(x - 2)/(x - 1), which simplifies to (x - 2) after canceling the common factor (x - 1). Substituting x = 1 gives 1 - 2 = -1. A common error is incorrectly factoring the quadratic or making substitution mistakes. The transferable technique is to factor quadratics systematically, cancel common factors that create indeterminate forms, and evaluate carefully.

Question 3

Evaluate lim⁡x→11−xx−1\lim\limits_{x\to 1}\dfrac{1-x}{\sqrt{x}-1}x→1lim​x​−11−x​ by simplifying the indeterminate form.

  1. 000
  2. −2-2−2 (correct answer)
  3. 222
  4. −1-1−1
  5. 111

Explanation: This limit creates 0/0, requiring rationalization of the denominator since it contains a radical. Multiply both numerator and denominator by the conjugate (√x + 1). The denominator becomes (√x - 1)(√x + 1) = x - 1. The expression becomes (1 - x)(√x + 1)/(x - 1) = -(x - 1)(√x + 1)/(x - 1) = -(√x + 1) after canceling (x - 1). Substituting x = 1 gives -(√1 + 1) = -(1 + 1) = -2. Students often struggle with rationalizing denominators or handling the negative sign properly. The technique is to rationalize denominators with conjugates and track signs carefully.

Question 4

Evaluate lim⁡x→1x3−1x−1\lim\limits_{x\to 1}\dfrac{x^3-1}{x-1}x→1lim​x−1x3−1​ by simplifying the expression.

  1. 111
  2. 000
  3. 222
  4. 333 (correct answer)
  5. −3-3−3

Explanation: This limit creates 0/0, requiring factorization using the difference of cubes formula. The numerator x³ - 1 factors as (x - 1)(x² + x + 1) using the pattern a³ - b³ = (a - b)(a² + ab + b²). The expression becomes (x - 1)(x² + x + 1)/(x - 1), which simplifies to x² + x + 1 after canceling (x - 1). Substituting x = 1 gives 1² + 1 + 1 = 3. Students often don't recognize the difference of cubes pattern or make errors applying the formula. The key technique is to identify cubic patterns and apply the appropriate factorization formulas systematically.

Question 5

Determine lim⁡x→1x+3−2x−1\lim\limits_{x\to 1}\dfrac{\sqrt{x+3}-2}{x-1}x→1lim​x−1x+3​−2​ by rationalizing the numerator.

  1. 14\dfrac{1}{4}41​ (correct answer)
  2. 12\dfrac{1}{2}21​
  3. 444
  4. 000
  5. −14-\dfrac{1}{4}−41​

Explanation: This limit involves a radical expression creating 0/0, requiring rationalization of the numerator. Multiply both numerator and denominator by the conjugate (√(x+3) + 2). The numerator becomes (√(x+3) - 2)(√(x+3) + 2) = (x+3) - 4 = x - 1. The expression becomes (x - 1)/((x - 1)(√(x+3) + 2)), which simplifies to 1/(√(x+3) + 2) after canceling (x - 1). Substituting x = 1 gives 1/(√(1+3) + 2) = 1/(√4 + 2) = 1/(2 + 2) = 1/4. Students often struggle with conjugate multiplication or make algebraic errors. The systematic approach for radical limits is to rationalize using conjugates and simplify step by step.

Question 6

Evaluate lim⁡x→0(1+x)3−1x\lim\limits_{x\to 0}\dfrac{(1+x)^3-1}{x}x→0lim​x(1+x)3−1​ using a single algebraic expansion/factorization step.

  1. 000
  2. 111
  3. 333 (correct answer)
  4. 666
  5. 999

Explanation: This limit creates 0/0, requiring expansion of the cubic expression in the numerator. Expand (1+x)³ using the binomial theorem: (1+x)³ = 1 + 3x + 3x² + x³. The numerator becomes 1 + 3x + 3x² + x³ - 1 = 3x + 3x² + x³ = x(3 + 3x + x²). The expression becomes x(3 + 3x + x²)/x, which simplifies to (3 + 3x + x²) after canceling x. Substituting x = 0 gives 3 + 0 + 0 = 3. A common error is incorrectly expanding the cubic or not factoring out x properly. The technique is to expand binomial expressions, factor common terms, cancel, and substitute.

Question 7

A temperature change uses (x−1)2x−1\dfrac{(x-1)^2}{x-1}x−1(x−1)2​; find lim⁡x→1(x−1)2x−1\lim\limits_{x\to 1}\dfrac{(x-1)^2}{x-1}x→1lim​x−1(x−1)2​.

  1. 222
  2. 111
  3. 000 (correct answer)
  4. −1-1−1
  5. −2-2−2

Explanation: This limit involves a perfect square expression that creates 0/0, requiring careful algebraic simplification. The numerator (x-1)² can be written as (x-1)(x-1). The expression becomes (x-1)(x-1)/(x-1), which simplifies to (x-1) after canceling one factor of (x-1). Substituting x = 1 gives 1 - 1 = 0. Students often incorrectly cancel both factors or fail to recognize the perfect square structure. The key strategy is to expand or factor perfect squares appropriately, cancel only one instance of common factors, and substitute carefully.

Question 8

Compute lim⁡x→5x2−25x−5\lim\limits_{x\to 5}\dfrac{x^2-25}{x-5}x→5lim​x−5x2−25​, an indeterminate form that simplifies by factoring.

  1. 252525
  2. 101010 (correct answer)
  3. 555
  4. 000
  5. −10-10−10

Explanation: This limit involves the indeterminate form 0/0, requiring factorization of the difference of squares. The numerator x² - 25 factors as (x - 5)(x + 5). The expression becomes (x - 5)(x + 5)/(x - 5), which simplifies to (x + 5) after canceling the common factor (x - 5). Substituting x = 5 gives 5 + 5 = 10. A frequent error is not recognizing the difference of squares pattern or incorrectly factoring. The systematic strategy is to identify special polynomial forms like difference of squares, factor completely, cancel common terms, and then evaluate by substitution.

Question 9

Determine lim⁡x→1x2+x−2x−1\lim\limits_{x\to 1}\dfrac{x^2+x-2}{x-1}x→1lim​x−1x2+x−2​, which initially gives 00\frac{0}{0}00​.

  1. 111
  2. 000
  3. −1-1−1
  4. 222
  5. 333 (correct answer)

Explanation: This limit initially gives 00\frac{0}{0}00​, requiring factorization of the numerator. The quadratic x2+x−2x^2 + x - 2x2+x−2 factors as (x+2)(x−1)(x + 2)(x - 1)(x+2)(x−1) by finding two numbers that multiply to −2-2−2 and add to 111. The expression becomes (x+2)(x−1)x−1\frac{(x + 2)(x - 1)}{x - 1}x−1(x+2)(x−1)​, which simplifies to (x+2)(x + 2)(x+2) after canceling the common factor (x−1)(x - 1)(x−1). Substituting x=1x = 1x=1 gives 1+2=31 + 2 = 31+2=3. Students often struggle with factoring quadratics or make sign errors. The key approach is to factor the numerator completely, identify common factors with the denominator, cancel appropriately, and substitute.

Question 10

For a rate near x=0x=0x=0, find lim⁡x→01+x−1x\lim\limits_{x\to 0}\dfrac{\sqrt{1+x}-1}{x}x→0lim​x1+x​−1​.

  1. 111
  2. 12\dfrac{1}{2}21​ (correct answer)
  3. 000
  4. 222
  5. −12-\dfrac{1}{2}−21​

Explanation: This limit involves a radical expression creating 0/0, requiring rationalization of the numerator. Multiply both numerator and denominator by the conjugate (1+x+1)(\sqrt{1+x} + 1)(1+x​+1). The numerator becomes (1+x−1)(1+x+1)=(1+x)−1=x(\sqrt{1+x} - 1)(\sqrt{1+x} + 1) = (1+x) - 1 = x(1+x​−1)(1+x​+1)=(1+x)−1=x. The expression simplifies to xx(1+x+1)=11+x+1\frac{x}{x(\sqrt{1+x} + 1)} = \frac{1}{\sqrt{1+x} + 1}x(1+x​+1)x​=1+x​+11​ after canceling x. Substituting x = 0 gives 11+1=11+1=12\frac{1}{\sqrt{1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}1​+11​=1+11​=21​. Students often forget to use the conjugate or make errors during rationalization. The essential technique for radical limits is to multiply by the conjugate to eliminate radicals and create factorable forms.

Question 11

Compute lim⁡x→−1x2−1x+1\lim\limits_{x\to -1}\dfrac{x^2-1}{x+1}x→−1lim​x+1x2−1​ by simplifying the expression first.

  1. −2-2−2 (correct answer)
  2. 222
  3. −1-1−1
  4. 000
  5. 111

Explanation: This limit creates 0/0, requiring factorization of the difference of squares. The numerator x² - 1 factors as (x - 1)(x + 1). The expression becomes (x - 1)(x + 1)/(x + 1), which simplifies to (x - 1) after canceling the common factor (x + 1). Substituting x = -1 gives -1 - 1 = -2. A common error is not recognizing the difference of squares pattern or incorrectly canceling factors. The transferable method is to identify special polynomial forms, factor completely, cancel common terms that create indeterminate forms, and evaluate by substitution.

Question 12

Evaluate lim⁡x→04+x−2x\lim\limits_{x\to 0}\dfrac{\sqrt{4+x}-2}{x}x→0lim​x4+x​−2​ by rationalizing the numerator.

  1. 12\dfrac{1}{2}21​
  2. 14\dfrac{1}{4}41​ (correct answer)
  3. 444
  4. 000
  5. −14-\dfrac{1}{4}−41​

Explanation: This limit involves a radical creating 0/0, requiring rationalization using the conjugate. Multiply both numerator and denominator by (√(4+x) + 2). The numerator becomes (√(4+x) - 2)(√(4+x) + 2) = (4+x) - 4 = x. The expression simplifies to x/(x(√(4+x) + 2)) = 1/(√(4+x) + 2) after canceling x. Substituting x = 0 gives 1/(√4 + 2) = 1/(2 + 2) = 1/4. A common error is not recognizing the need for rationalization or making algebraic mistakes. The strategy for radical expressions is to rationalize using conjugates, simplify algebraically, and then evaluate by substitution.

Question 13

Compute lim⁡x→−2x2+5x+6x+2\lim\limits_{x\to -2}\dfrac{x^2+5x+6}{x+2}x→−2lim​x+2x2+5x+6​, which gives an indeterminate form before simplifying.

  1. −2-2−2
  2. 000
  3. 111
  4. −1-1−1 (correct answer)
  5. 222

Explanation: This limit creates the indeterminate form 0/00/00/0, necessitating algebraic manipulation before evaluation. The numerator x2+5x+6x^2 + 5x + 6x2+5x+6 factors as (x+2)(x+3)(x + 2)(x + 3)(x+2)(x+3) using standard quadratic factoring techniques. The expression becomes (x+2)(x+3)/(x+2)(x + 2)(x + 3)/(x + 2)(x+2)(x+3)/(x+2), which simplifies to (x+3)(x + 3)(x+3) after canceling the common factor (x+2)(x + 2)(x+2). Substituting x = -2 gives −2+3=−1-2 + 3 = -1−2+3=−1. Students often make errors by incorrectly factoring the quadratic or failing to identify the common factor. The systematic strategy is to factor the numerator completely, identify common factors with the denominator, cancel them, and then substitute.

Question 14

A geometric approximation uses lim⁡x→11−x21−x\lim_{x\to 1}\frac{1-x^2}{1-x}limx→1​1−x1−x2​. What is the value of the limit?

  1. 000
  2. −2-2−2
  3. 222 (correct answer)
  4. 111
  5. DNE\text{DNE}DNE

Explanation: This limit gives 0/0 when substituting x = 1, requiring algebraic manipulation. The numerator 1 - x² can be rewritten as -(x² - 1) = -(x - 1)(x + 1). The denominator 1 - x can be rewritten as -(x - 1). Our limit becomes lim[x→1] [-(x - 1)(x + 1)]/[-(x - 1)] = lim[x→1] (x + 1) after canceling -(x - 1) from both numerator and denominator. Substituting x = 1 gives 1 + 1 = 2. A common error is not recognizing that 1 - x² = -(x² - 1) or forgetting to handle the negative signs properly when factoring. When you see expressions like 1 - x² or a - b, consider rewriting them as -(x² - 1) or -(b - a) to reveal familiar factoring patterns.

Question 15

Evaluate lim⁡x→−1x2−1x+1\displaystyle\lim_{x\to -1}\frac{x^2-1}{x+1}x→−1lim​x+1x2−1​ by simplifying the expression first.

  1. −2-2−2 (correct answer)
  2. 000
  3. 222
  4. −1-1−1
  5. Does not exist

Explanation: This limit has the form 0/0 when x = -1, so we factor the numerator. The expression x² - 1 is a difference of squares: (x + 1)(x - 1). Our limit becomes lim[x→-1] [(x + 1)(x - 1)]/(x + 1), and we can cancel the (x + 1) terms since x approaches but doesn't equal -1. This simplifies to lim[x→-1] (x - 1) = -1 - 1 = -2. A common error is factoring x² - 1 as (x - 1)² instead of recognizing the difference of squares pattern. Always check: a² - b² = (a + b)(a - b), then cancel common factors before substituting.

Question 16

A rate is defined by x+5−3x−4\frac{\sqrt{x+5}-3}{x-4}x−4x+5​−3​; find lim⁡x→4x+5−3x−4\displaystyle\lim_{x\to 4}\frac{\sqrt{x+5}-3}{x-4}x→4lim​x−4x+5​−3​.

  1. 16\frac{1}{6}61​ (correct answer)
  2. 13\frac{1}{3}31​
  3. 19\frac{1}{9}91​
  4. 12\frac{1}{2}21​
  5. 112\frac{1}{12}121​

Explanation: This limit involves a radical expression creating 0/0 when x = 4. To resolve this, we multiply by the conjugate (√(x+5) + 3) over itself to rationalize. This gives us [(√(x+5) - 3)(√(x+5) + 3)]/[(x-4)(√(x+5) + 3)] = (x+5-9)/[(x-4)(√(x+5) + 3)] = (x-4)/[(x-4)(√(x+5) + 3)]. After canceling (x-4), we get 1/(√(x+5) + 3), and substituting x = 4 yields 1/(√9 + 3) = 1/(3 + 3) = 1/6. A common error is forgetting to multiply both numerator and denominator by the conjugate. The key strategy: when radicals create 0/0, use conjugate multiplication to eliminate the radical difference.

Question 17

A model uses x+9−3x\dfrac{\sqrt{x+9}-3}{x}xx+9​−3​; determine lim⁡x→0x+9−3x\lim\limits_{x\to 0}\dfrac{\sqrt{x+9}-3}{x}x→0lim​xx+9​−3​.

  1. 16\dfrac{1}{6}61​ (correct answer)
  2. 13\dfrac{1}{3}31​
  3. 19\dfrac{1}{9}91​
  4. 000
  5. 666

Explanation: This limit involves a radical expression that creates 0/0, requiring rationalization of the numerator. Multiply both numerator and denominator by the conjugate (x+9+3\sqrt{x+9} + 3x+9​+3). The numerator becomes (x+9−3\sqrt{x+9} - 3x+9​−3)(x+9+3\sqrt{x+9} + 3x+9​+3) = (x+9) - 9 = x. The expression simplifies to xx(x+9+3)\frac{x}{x(\sqrt{x+9} + 3)}x(x+9​+3)x​ = 1x+9+3\frac{1}{\sqrt{x+9} + 3}x+9​+31​ after canceling x. Substituting x = 0 gives 19+3\frac{1}{\sqrt{9} + 3}9​+31​ = 13+3\frac{1}{3 + 3}3+31​ = 16\frac{1}{6}61​. Students often forget to rationalize or make algebraic errors during the process. The key strategy for radical limits is to rationalize using the conjugate to eliminate the radical from the numerator.

Question 18

Compute lim⁡x→0(3+x)2−9x\lim\limits_{x\to 0}\dfrac{(3+x)^2-9}{x}x→0lim​x(3+x)2−9​ by simplifying the expression.

  1. 333
  2. 666 (correct answer)
  3. 999
  4. 000
  5. −6-6−6

Explanation: This limit creates 0/0, requiring expansion and simplification of the squared binomial. Expand (3+x)²: (3+x)² = 9 + 6x + x². The numerator becomes 9 + 6x + x² - 9 = 6x + x² = x(6 + x). The expression becomes x(6 + x)/x, which simplifies to (6 + x) after canceling x. Substituting x = 0 gives 6 + 0 = 6. Students often make errors expanding binomial squares or not factoring out x correctly. The strategy is to expand carefully, factor common terms, cancel appropriately, and substitute the limit value.

Question 19

Determine lim⁡x→4x2−16x2−8x+16\lim\limits_{x\to 4}\dfrac{x^2-16}{x^2-8x+16}x→4lim​x2−8x+16x2−16​ using one factoring step.

  1. 000
  2. 111
  3. 222
  4. Does not exist (correct answer)
  5. 888

Explanation: This limit involves factoring both numerator and denominator, creating a more complex 0/0 form. The numerator x² - 16 factors as (x - 4)(x + 4), and the denominator x² - 8x + 16 factors as (x - 4)². The expression becomes (x - 4)(x + 4)/(x - 4)², which simplifies to (x + 4)/(x - 4) after canceling one (x - 4). As x approaches 4, the numerator approaches 8 while the denominator approaches 0, making the limit undefined (does not exist). Students often miss that one factor remains in the denominator. The key is to factor completely and recognize when limits don't exist due to division by zero.

Question 20

A function is defined by f(x)=x+9−3xf(x)=\frac{\sqrt{x+9}-3}{x}f(x)=xx+9​−3​ for x≠0x\ne0x=0; find lim⁡x→0f(x)\lim_{x\to0}f(x)limx→0​f(x).

  1. 13\frac{1}{3}31​
  2. 16\frac{1}{6}61​ (correct answer)
  3. 19\frac{1}{9}91​
  4. 000
  5. 32\frac{3}{2}23​

Explanation: This limit has the indeterminate form 0/0, requiring rationalization since we have a square root. We multiply both numerator and denominator by the conjugate (√(x + 9) + 3). The numerator becomes (√(x + 9) - 3)(√(x + 9) + 3) = (x + 9) - 9 = x using the difference of squares. The denominator becomes x(√(x + 9) + 3), so our expression simplifies to x/[x(√(x + 9) + 3)] = 1/(√(x + 9) + 3). Substituting x = 0 gives 1/(√9 + 3) = 1/(3 + 3) = 1/6. A common error is forgetting to multiply the denominator by the conjugate or making arithmetic mistakes. The key strategy for limits with square roots is to rationalize using conjugates.