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AP Calculus BC Practice Quiz: Determining Limits Using Algebraic Manipulation

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

Evaluate the limit: lim (x→2) (x² - 4) / (x - 2)

All Questions (7)

Evaluate the limit: lim (x→2) (x² - 4) / (x - 2)

A) 0

B) 2

C) 4

D) The limit does not exist.

Correct Answer: C

Direct substitution of x=2 results in the indeterminate form 0/0. To find the limit, we must first rearrange the expression into an equivalent form. We can factor the numerator as a difference of squares: x² - 4 = (x - 2)(x + 2). The expression becomes [(x - 2)(x + 2)] / (x - 2). We can cancel the (x - 2) term, leaving the equivalent expression x + 2. Now, we can evaluate the limit by substituting x=2 into the simplified expression: 2 + 2 = 4.

Find the limit: lim (x→0) (√(x + 9) - 3) / x

A) 1/6

B) 1/3

C) 0

D) The limit does not exist.

Correct Answer: A

Direct substitution of x=0 yields the indeterminate form 0/0. To evaluate the limit, we can use an equivalent expression obtained by multiplying the numerator and denominator by the conjugate of the numerator, which is √(x + 9) + 3. This gives [ (√(x + 9) - 3)(√(x + 9) + 3) ] / [ x(√(x + 9) + 3) ]. The numerator simplifies to (x + 9) - 9 = x. The expression becomes x / [ x(√(x + 9) + 3) ]. Canceling the x term leaves 1 / (√(x + 9) + 3). Now, substituting x=0 gives 1 / (√(0 + 9) + 3) = 1 / (3 + 3) = 1/6.

Evaluate the limit: lim (h→0) [ (1/(4+h)) - (1/4) ] / h

A) 1/16

B) -1/16

C) 1/4

D) -1/4

Correct Answer: B

Direct substitution of h=0 results in 0/0. To find the limit, we need to simplify the complex fraction. First, find a common denominator for the terms in the numerator: [ 4 - (4+h) ] / [ 4(4+h) ] = -h / [ 4(4+h) ]. The entire expression is now ( -h / [4(4+h)] ) / h. We can cancel h from the numerator and denominator, which results in the equivalent expression -1 / [ 4(4+h) ]. Evaluating the limit by substituting h=0 gives -1 / [ 4(4+0) ] = -1/16.

Let f, g, and h be functions satisfying h(x) ≤ f(x) ≤ g(x) for all x near c, except possibly at c. If lim (x→c) h(x) = 5 and lim (x→c) g(x) = 5, what is lim (x→c) f(x)?

A) 0

B) 5

C) The limit cannot be determined from the information given.

D) The limit does not exist.

Correct Answer: B

This is a direct application of the Squeeze Theorem. The theorem states that if a function f(x) is bounded between two other functions, h(x) and g(x), and both h(x) and g(x) approach the same limit L as x approaches c, then f(x) must also approach that same limit L. Since both the lower bound h(x) and the upper bound g(x) approach the limit 5, the function f(x) is 'squeezed' and must also approach the limit 5.

What is the value of lim (x→-3) (x² + x - 6) / (x² - 9)?

A) 5/6

B) -1/6

C) 0

D) The limit does not exist.

Correct Answer: A

Substituting x=-3 into the expression gives the indeterminate form 0/0. We must rearrange the expression by factoring the numerator and the denominator. The numerator factors to (x+3)(x-2). The denominator factors to (x-3)(x+3). The expression becomes [(x+3)(x-2)] / [(x-3)(x+3)]. After canceling the common factor of (x+3), we get the equivalent expression (x-2)/(x-3). Now, we can substitute x=-3: (-3-2)/(-3-3) = -5/-6 = 5/6.

Evaluate the limit: lim (x→1) (x³ - 1) / (x - 1)

A) 0

B) 1

C) 3

D) The limit does not exist.

Correct Answer: C

Direct substitution of x=1 results in the indeterminate form 0/0. To find the limit, we can create an equivalent expression by factoring the numerator, which is a difference of cubes: x³ - 1 = (x - 1)(x² + x + 1). The expression becomes [(x - 1)(x² + x + 1)] / (x - 1). We can cancel the (x - 1) term, leaving the simplified expression x² + x + 1. Evaluating the limit by substituting x=1 gives 1² + 1 + 1 = 3.

For a function f(x), it is known that (4x - 2) ≤ f(x) ≤ (x² + 2) for x ≥ 0. Which of the following is the value of lim (x→2) f(x)?

A) 0

B) 6

C) 8

D) The limit cannot be determined.

Correct Answer: B

This problem requires the use of the Squeeze Theorem. We are given that f(x) is bounded by h(x) = 4x - 2 and g(x) = x² + 2. To find the limit of f(x) as x approaches 2, we first find the limits of the bounding functions. lim (x→2) h(x) = lim (x→2) (4x - 2) = 4(2) - 2 = 6. And lim (x→2) g(x) = lim (x→2) (x² + 2) = 2² + 2 = 6. Since the limits of both the lower and upper bounding functions are equal to 6, the Squeeze Theorem states that the limit of f(x) as x approaches 2 must also be 6.