AP Calculus AB: Evaluating Limits Algebraically (Drill 1)

Question 1. What is \( \lim_{x \to 2} \dfrac{x^3 - 3x + 5}{x^2 + 1} \)?

• A) \( 1 \)
• B) \( \dfrac{13}{5} \)
• C) \( \dfrac{7}{3} \)
• D) \( \dfrac{7}{5} \) ✓

Explanation: Choice D is correct. The denominator at \( x = 2 \) is \( 4 + 1 = 5 \neq 0 \), so direct substitution applies. Numerator \( = (2)^3 - 3(2) + 5 = 8 - 6 + 5 = 7 \); denominator \( = 5 \). The limit equals \( \dfrac{7}{5} \). Choice A is off because evaluating only the constant term in the numerator gives \( \dfrac{5}{5} = 1 \). Choice B doesn't fit: dropping the \( -3x \) term entirely gives numerator \( = 8 + 5 = 13 \), and dividing by the correct denominator of 5 gives \( \dfrac{13}{5} \). Choice C is wrong because substituting into the denominator as \( 2 + 1 = 3 \) (forgetting to square) gives \( \dfrac{7}{3} \).

Question 2. What is \( \lim_{x \to 4} \dfrac{x^2 - 16}{x - 4} \)?

• A) \( 0 \)
• B) \( 4 \)
• C) \( 8 \) ✓
• D) The limit does not exist.

Explanation: Choice C is correct. Direct substitution yields \( \dfrac{0}{0} \), an indeterminate form. Factor the numerator: \( x^2 - 16 = (x-4)(x+4) \). Cancel \( (x-4) \): \( \lim_{x \to 4}(x+4) = 8 \). Choice A is off because seeing the numerator become 0 and concluding the limit is 0, ignoring that the denominator is also 0. Choice B doesn't fit: incorrectly concluding the limit equals the x-value being approached, \( x = 4 \). Choice D is wrong because \( \dfrac{0}{0} \) is indeterminate, not undefined, after factoring, the limit exists and equals 8.

Question 3. What is \( \lim_{x \to 1} \dfrac{x^3 - 1}{x - 1} \)?

• A) \( 3 \) ✓
• B) \( 2 \)
• C) \( 0 \)
• D) The limit does not exist.

Explanation: Choice A is correct. Direct substitution gives \( \dfrac{0}{0} \). Apply the difference of cubes factorization: \( x^3 - 1 = (x-1)(x^2+x+1) \). Cancel \( (x-1) \): \( \lim_{x \to 1}(x^2+x+1) = 1 + 1 + 1 = 3 \). Choice B is off because misremembering the difference of cubes formula as \( (x-1)(x^2+1) \) and evaluating \( (1^2+1) = 2 \). Choice C doesn't fit: seeing \( \dfrac{0}{0} \) and reporting the numerator's value as the limit. Choice D is wrong because failing to attempt algebraic simplification before concluding the limit does not exist.

Question 4. What is \( \lim_{x \to 9} \dfrac{\sqrt{x} - 3}{x - 9} \)?

• A) \( 0 \)
• B) \( 1 \)
• C) \( 3 \)
• D) \( \dfrac{1}{6} \) ✓

Explanation: Choice D is correct. Direct substitution gives \( \dfrac{0}{0} \). Multiply numerator and denominator by the conjugate \( (\sqrt{x}+3) \): the numerator becomes \( (\sqrt{x}-3)(\sqrt{x}+3) = x - 9 \); the denominator becomes \( (x-9)(\sqrt{x}+3) \). Cancel \( (x-9) \): \( \dfrac{1}{\sqrt{x}+3} \). As \( x \to 9 \): \( \dfrac{1}{\sqrt{9}+3} = \dfrac{1}{6} \). Choice A doesn't work because canceling and then failing to evaluate the remaining expression, reporting 0. Choice B falls short because forgetting to evaluate \( \sqrt{x}+3 \) at \( x = 9 \) and writing \( \dfrac{1}{1} = 1 \). Choice C misses the mark: evaluating only \( \sqrt{9} = 3 \) and reporting it as the limit.

Question 5. A student evaluates \( \lim_{x \to 5} \dfrac{x^2 - 25}{x - 5} \) and writes: “Since substituting \( x = 5 \) gives \( \dfrac{0}{0} \), the limit does not exist.” Which of the following best describes the student’s error?

• A) The student should have used L’Hôpital’s Rule, which gives the correct answer of 0.
• B) The student correctly identified the form but should have reported the limit as 0.
• C) The \( \dfrac{0}{0} \) form is indeterminate, not undefined, algebraic simplification reveals the limit equals 10. ✓
• D) The limit truly does not exist because the function is undefined at \( x = 5 \).

Explanation: Choice C is correct. \( \dfrac{0}{0} \) is an indeterminate form, not a definitive sign that no limit exists. Factoring: \( \dfrac{x^2-25}{x-5} = \dfrac{(x-5)(x+5)}{x-5} = x+5 \) for \( x \neq 5 \). As \( x \to 5 \): \( x + 5 = 10 \). Choice A is wrong because L’Hôpital’s Rule is not the expected Unit 1 technique; also, applying it correctly gives \( 2x \) evaluated at \( x = 5 \), which equals 10, not 0. Choice B doesn't work because the limit is 10, not 0. Choice D falls short because a function can be undefined at a point yet still have a limit there, the limit describes behavior as \( x \) approaches the point, not the value at it.