Drill 1 · Math · Linearization and L'Hôpital's Rule
AP Calculus AB: Linearization and L’Hôpital’s Rule (Drill 1) is a Math practice drill covering Linearization and L'Hôpital's Rule. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Practice using local linear approximation to estimate function values and applying L'Hôpital's Rule to evaluate limits in indeterminate forms. These AP Calculus AB skills appear in both multiple-choice and free-response sections of the exam.
Question 1. Let \( f(x) = \sqrt{x} \). Which of the following is the linearization of \( f \) at \( x = 9 \)?
Explanation: Choice A is correct. The linearization formula is \( L(x) = f(a) + f'(a)(x - a) \). Here \( f(9) = 3 \) and \( f'(x) = \dfrac{1}{2\sqrt{x}} \), so \( f'(9) = \dfrac{1}{6} \). Thus \( L(x) = 3 + \dfrac{1}{6}(x - 9) \). Choice B is incorrect because it uses \( \dfrac{1}{3} \) as the slope, the student forgot the 2 in the denominator of \( f'(x) = \dfrac{1}{2\sqrt{x}} \), computing \( \dfrac{1}{\sqrt{9}} = \dfrac{1}{3} \) instead of \( \dfrac{1}{2\sqrt{9}} = \dfrac{1}{6} \). Choice C is incorrect because the linearization must be centered at \( a = 9 \), not \( a = 3 \); the student swapped \( f(a) \) and \( a \) in the formula. Choice D is incorrect because the student wrote \( f'(x) = \dfrac{1}{2} \) without evaluating \( \dfrac{1}{2\sqrt{x}} \) at \( x = 9 \), leaving the slope as \( \dfrac{1}{2} \) rather than \( \dfrac{1}{6} \).
Question 2. What is \( \displaystyle\lim_{x \to 0} \dfrac{\sin 3x}{x} \)?
Explanation: Choice C is correct. As \( x \to 0 \), both numerator and denominator approach 0, giving the indeterminate form \( \dfrac{0}{0} \). Applying L'Hôpital's Rule, differentiate numerator and denominator: \( \dfrac{d}{dx}(\sin 3x) = 3\cos 3x \) and \( \dfrac{d}{dx}(x) = 1 \). The limit becomes \( \displaystyle\lim_{x \to 0} \dfrac{3\cos 3x}{1} = 3\cos(0) = 3 \). Choice A is incorrect because the student observed the numerator approaching 0 without recognizing that the denominator also approaches 0, making this an indeterminate form that cannot be evaluated by direct substitution alone. Choice B is incorrect because the student applied the standard result \( \displaystyle\lim_{x \to 0} \dfrac{\sin x}{x} = 1 \) without accounting for the factor of 3 inside the sine function. Choice D is incorrect because the indeterminate form \( \dfrac{0}{0} \) resolves to a finite value of 3; the limit exists.
Question 3. Let \( g \) be a twice-differentiable function with \( g(2) = 5 \), \( g'(2) = -1 \), and \( g''(x) > 0 \) for all \( x \). Which of the following best describes the linearization \( L(x) \) of \( g \) at \( x = 2 \) used to approximate \( g(2.1) \)?
Explanation: Choice B is correct. When \( g''(x) > 0 \) for all \( x \), the graph of \( g \) is concave up everywhere. A concave up graph curves upward away from any tangent line, which means the tangent line lies below the curve. The linearization \( L(x) \) is the tangent line at \( x = 2 \), so \( L(2.1) < g(2.1) \), the linear approximation is an underestimate. Choice A is incorrect because it correctly identifies the concavity but draws the wrong conclusion: concave up means the tangent line lies below the curve, producing an underestimate, not an overestimate. Choice C is incorrect because whether \( g \) is increasing or decreasing (determined by the sign of \( g'(2) = -1 \)) does not determine whether the linearization over- or underestimates; that is determined solely by concavity. Choice D is incorrect for the same reason as Choice C: the fact that \( g \) is decreasing at \( x = 2 \) is irrelevant to whether \( L(x) \) is an over- or underestimate.
Question 4. What is \( \displaystyle\lim_{x \to \infty} \dfrac{3x^2 + 1}{5x^2 - 2x} \)?
Explanation: Choice B is correct. As \( x \to \infty \), both numerator and denominator approach \( \infty \), giving the indeterminate form \( \dfrac{\infty}{\infty} \). Applying L'Hôpital's Rule: \( \dfrac{6x}{10x - 2} \), still \( \dfrac{\infty}{\infty} \). Applying L'Hôpital's Rule again: \( \dfrac{6}{10} = \dfrac{3}{5} \). Equivalently, dividing numerator and denominator by \( x^2 \) gives \( \dfrac{3 + 1/x^2}{5 - 2/x} \to \dfrac{3}{5} \) as \( x \to \infty \). Choice A is incorrect because a limit of 0 at infinity would require the denominator to have strictly higher degree than the numerator; here both have degree 2. Choice C is incorrect because the student compared only the constant terms, dividing the 1 in the numerator by the 5 in the denominator while ignoring the dominant \( x^2 \) terms entirely. Choice D is incorrect because a limit of \( \infty \) would require the numerator to have strictly higher degree than the denominator; here the degrees are equal.
Question 5. A student evaluates \( \displaystyle\lim_{x \to 0} \dfrac{x + 2}{x} \) using L'Hôpital's Rule, differentiating numerator and denominator to obtain a limit of 1. Which of the following best identifies the error?
Explanation: Choice B is correct. L'Hôpital's Rule may only be applied when the limit produces the indeterminate form \( \dfrac{0}{0} \) or \( \dfrac{\infty}{\infty} \). Here, as \( x \to 0 \), the numerator \( x + 2 \to 2 \) while the denominator \( x \to 0 \), giving the form \( \dfrac{2}{0} \), not an indeterminate form. As \( x \to 0^{+} \) the expression \( \to +\infty \) and as \( x \to 0^{-} \) it \( \to -\infty \), so the two-sided limit does not exist. The student's error was applying the rule without first verifying that the limit produces an indeterminate form. Choice A is incorrect because the derivative of \( x + 2 \) is indeed 1, not 0; the differentiation was performed correctly. The problem is not with the calculus but with the invalid application of the rule. Choice C is incorrect because repeated application of an inapplicable rule cannot produce a correct result. Choice D is incorrect: the answer of 1 is wrong, and the method used to obtain it is invalid.