AP Calculus AB Quiz: Lhospitals Rule
20 questions · exam conditions
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Lhospitals RuleQuestion 1 of 20

Find lim⁡x→0sin⁡(x2)x2\displaystyle \lim_{x\to 0}\frac{\sin(x^2)}{x^2}.

00
22
11
−1-1
12\dfrac{1}{2}
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AP Calculus AB Quiz

AP Calculus AB Quiz: Lhospitals Rule

Practice Lhospitals Rule in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Lhospitals Rule, 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

Find lim⁡x→0sin⁡(x2)x2\displaystyle \lim_{x\to 0}\frac{\sin(x^2)}{x^2}.

  1. 00
  2. 22
  3. 11 (correct answer)
  4. −1-1
  5. 12\dfrac{1}{2}

Explanation: 0/00/0 form; L'Hôpital: 2xcos⁡(x2)2x=cos⁡(x2)→1\frac{2x \cos(x^2)}{2x} = \cos(x^2) \to 1. Or sin⁡(u)u→1\frac{\sin(u)}{u} \to 1, u=x2u = x^2, adjusted. Mistake: thinking it's sin⁡x/x\sin x / x. Another: substitution error. Strategy: variable substitution for inner functions.

Question 2

Compute lim⁡x→0tan⁡(x2)x2\displaystyle \lim_{x\to 0}\frac{\tan(x^2)}{x^2}.

  1. 11 (correct answer)
  2. 00
  3. 22
  4. 12\dfrac{1}{2}
  5. −1-1

Explanation: 0/0; L'Hôpital: (2x sec²(x²))/(2x) = sec²(x²) →1. Similar to tan u /u →1. Wrong: confusing with tan x /x. Error: incorrect chain rule. Strategy: apply known limits to composites.

Question 3

A model uses lim⁡x→0e4x−1−4xx2\lim_{x\to0}\frac{e^{4x}-1-4x}{x^2}. What is the value of the limit?

  1. 44
  2. 88 (correct answer)
  3. 1616
  4. 00
  5. 3232

Explanation: This limit has the indeterminate form 0/00/0 since e4x−1−4xe^{4x} - 1 - 4x approaches 0 as x approaches 0 (the linear approximation of e4xe^{4x} near 0 is 1+4x1 + 4x). Applying L'Hôpital's Rule once gives (4e4x−4)/(2x)(4e^{4x} - 4)/(2x), which is still 0/00/0. Applying it again yields 16e4x/2=8e4x16e^{4x}/2 = 8e^{4x}, which equals 88 when x = 0. A tempting error is to think that e4x−1≈4xe^{4x} - 1 \approx 4x near zero, making the numerator approximately 0, but this ignores the quadratic term in the Taylor expansion. The strategy is to recognize that when exponential functions appear with their linear approximations subtracted, you often need L'Hôpital's Rule twice to resolve the indeterminacy.

Question 4

Using L'Hospital's Rule, find lim⁡x→∞xln⁡(x2+1)\lim_{x\to\infty}\frac{x}{\ln(x^2+1)}.

  1. 00
  2. 11
  3. 22
  4. ∞\infty (correct answer)
  5. 12\frac12

Explanation: As xx approaches infinity, both xx and ln⁡(x2+1)\ln(x^2 + 1) approach infinity, creating an ∞/∞\infty / \infty indeterminate form. Applying L'Hôpital's Rule gives 12xx2+1=x2+12x\frac{1}{\frac{2x}{x^2 + 1}} = \frac{x^2 + 1}{2x}. As xx approaches infinity, this simplifies to x22x=x2\frac{x^2}{2x} = \frac{x}{2}, which approaches infinity. The limit is therefore infinity, not a finite value. A common mistake is to think that since ln grows slowly, the limit might be finite, but x grows faster than ln⁡(x2+1)≈2ln⁡(x)\ln(x^2 + 1) \approx 2\ln(x). The strategy is to simplify after each application of L'Hôpital's Rule and check whether the result still approaches infinity.

Question 5

As t→0t\to0, the average velocity is sin⁡(5t)−5tt3\frac{\sin(5t)-5t}{t^3}. What is the limit?

  1. 00
  2. −1256-\frac{125}{6} (correct answer)
  3. 1256\frac{125}{6}
  4. −125-125
  5. 252\frac{25}{2}

Explanation: As t approaches 0, both the numerator sin(5t) - 5t and denominator t³ approach 0, creating a 0/0 indeterminate form that requires L'Hôpital's Rule. Applying L'Hôpital's Rule once gives (5cos(5t) - 5)/(3t²), which is still 0/0 at t = 0. Applying it again yields (-25sin(5t))/(6t), still 0/0. A third application gives (-125cos(5t))/6, which evaluates to -125/6 when t = 0. A common mistake is to stop after one or two applications when the form is still indeterminate. The key strategy is to continue applying L'Hôpital's Rule until you get a determinate form, checking the limit after each differentiation.

Question 6

Compute lim⁡x→0ln⁡(1+x2)x2\displaystyle \lim_{x\to 0}\frac{\ln(1+x^2)}{x^2}.

  1. 00
  2. 22
  3. 11 (correct answer)
  4. 12\dfrac{1}{2}
  5. −1-1

Explanation: As x approaches 0, it's 0/0, so L'Hôpital's Rule: derivative (2x/(1+x²))/(2x) = 1/(1+x²) → 1. Known limit ln(1+u)/u → 1 with u = x². A common mistake is direct substitution giving 0/0 without applying the rule. Another tempting error is confusing with other logs. A transferable strategy is substituting variables for composition limits.

Question 7

As t→0t\to 0, what is lim⁡t→0e5t−1−5tt2\displaystyle \lim_{t\to 0}\frac{e^{5t}-1-5t}{t^2}?

  1. 00
  2. 252\dfrac{25}{2} (correct answer)
  3. 2525
  4. 55
  5. 52\dfrac{5}{2}

Explanation: The limit as t approaches 0 of (e5te^{5t} - 1 - 5t)/t^2 results in the indeterminate form 0/0, necessitating L'Hôpital's Rule. Differentiating the numerator gives 5e^{5t} - 5 and the denominator 2t, which still yields 0/0. Applying L'Hôpital's Rule again produces 25e^{5t} in the numerator and 2 in the denominator, evaluating to 25/2 at t=0. A tempting wrong approach is to mistakenly conclude the limit is 0 after the first differentiation without checking the form. Instead, recognize that multiple applications may be needed for higher-order terms. A transferable strategy is to apply L'Hôpital's Rule repeatedly until the indeterminate form is resolved, ensuring each step is verified.

Question 8

Compute lim⁡x→0ln⁡(1+3x)−3xx2\displaystyle \lim_{x\to 0}\frac{\ln(1+3x)-3x}{x^2}.

  1. −92-\dfrac{9}{2} (correct answer)
  2. 92\dfrac{9}{2}
  3. −9-9
  4. 00
  5. 33

Explanation: Evaluating lim (ln(1+3x) - 3x)/x^2 as x→0 gives 0/0, requiring L'Hôpital's Rule. Derivatives are 3/(1+3x) - 3 over 2x, still 0/0; second derivatives yield -9/(1+3x)^2 over 2, which is -9/2 at x=0. One might wrongly plug in after the first step, thinking it's resolved. The key is recognizing the need for two applications due to the quadratic behavior. Taylor series for ln(1+3x) confirms the -9/2 limit. A transferable strategy is to use L'Hôpital's Rule for logarithmic indeterminates, applying it multiple times and cross-checking with series expansions.

Question 9

Find lim⁡x→0sin⁡x−xcos⁡xx3\displaystyle \lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}.

  1. 00
  2. 12\dfrac{1}{2}
  3. 11
  4. −12-\dfrac{1}{2}
  5. 13\dfrac{1}{3} (correct answer)

Explanation: The limit (sin x - x cos x)/x^3 is 0/0, but one L'Hôpital gives (x sin x)/ (3x23x^2) = sin x /(3x) → 1/3. Wrongly applying more times complicates unnecessarily. The step simplifies after first differentiation to a known limit. Taylor confirms x^3/3. This shows not always needing max applications. A transferable strategy is checking if form resolves early.

Question 10

Compute lim⁡x→0e3x−1−3xx2\displaystyle \lim_{x\to 0}\frac{e^{3x}-1-3x}{x^2}.

  1. 92\dfrac{9}{2} (correct answer)
  2. 99
  3. 33
  4. 32\dfrac{3}{2}
  5. 00

Explanation: The limit (e3xe^{3x} - 1 - 3x)/x^2 is 0/0, L'Hôpital twice gives 9/2. Mistakenly halving incorrectly yields 9. Taylor: (9x29x^2)/2 term. The step identifies second-order. It matches 9/2. A transferable strategy is exponential series for quick limits.

Question 11

Evaluate lim⁡x→0ln⁡(1+x)−xx2\displaystyle \lim_{x\to 0}\frac{\ln(1+x)-x}{x^2}.

  1. 12\dfrac{1}{2}
  2. −12-\dfrac{1}{2} (correct answer)
  3. −1-1
  4. 00
  5. 11

Explanation: The limit (ln(1+x) - x)/x^2 as x→0 is 0/0, use L'Hôpital. Derivatives: 1/(1+x) - 1 / 2x, 0/0; second: -1/(1+x)^2 / 2 = -1/2. One might incorrectly assume it's 0 from direct substitution. The step involves Taylor: ln(1+x) ≈ x - x^2/2, yielding -1/2. This captures the quadratic term. A transferable strategy is applying L'Hôpital for logs and confirming with series for signs.

Question 12

Evaluate lim⁡x→01−cos⁡(x2)x4\displaystyle \lim_{x\to 0}\frac{1-\cos(x^2)}{x^4}.

  1. 00
  2. 12\dfrac{1}{2} (correct answer)
  3. 11
  4. 22
  5. 14\dfrac{1}{4}

Explanation: 0/00/0; twice L'Hôpital: derivative sin⁡(x2)⋅2x/4x3\sin(x^2) \cdot 2x / 4x^3, adjust; better identity 1−cos⁡u=u221 - \cos u = \frac{u^2}{2} approx, u=x2x^2, gives 12\frac{1}{2}. Mistake: using 1−cos⁡x1 - \cos x. Error: insufficient applications. Strategy: use trig identities before rule.

Question 13

To compare growth rates, evaluate lim⁡x→∞ln⁡xx\lim_{x\to\infty}\frac{\ln x}{\sqrt{x}} using L'Hospital's Rule.

  1. ∞\infty
  2. 11
  3. 00 (correct answer)
  4. 12\frac{1}{2}
  5. −∞-\infty

Explanation: As x approaches infinity, both ln⁡(x)\ln(x) and x\sqrt{x} approach infinity, giving the indeterminate form ∞/∞\infty / \infty. Applying L'Hôpital's Rule, we differentiate to get 1/x1/(2x)=2xx=2x\frac{1/x}{1/(2\sqrt{x})} = \frac{2\sqrt{x}}{x} = \frac{2}{\sqrt{x}}. As x approaches infinity, 2x\frac{2}{\sqrt{x}} approaches 0, showing that logarithmic growth is slower than square root growth. A common mistake is to think that since both functions grow without bound, the limit doesn't exist or equals 1. The key insight is that L'Hôpital's Rule helps compare growth rates: when the limit is 0, the numerator grows more slowly than the denominator.

Question 14

For x→0x\to 0, the efficiency model is e2x−1−2xx2\frac{e^{2x}-1-2x}{x^2}. What is the limit?

  1. 44
  2. 00
  3. 22 (correct answer)
  4. 11
  5. 12\dfrac{1}{2}

Explanation: This limit presents the indeterminate form 0/0 as both e^(2x) - 1 - 2x and x² approach 0 when x → 0. Applying L'Hôpital's Rule once gives (2e^(2x) - 2)/(2x), which simplifies to (e^(2x) - 1)/x, still 0/0. Applying L'Hôpital's Rule again yields 2e^(2x)/1 = 2e^(2x), which evaluates to 2 as x → 0. Students often make the error of forgetting to differentiate the constant term -2x in the numerator, leading to incorrect results. For exponential expressions of the form e^(ax) - 1 - ax, the limit as x → 0 of the expression divided by x² is always a²/2.

Question 15

Compute lim⁡x→0ex−1−sin⁡xx2\displaystyle \lim_{x\to 0}\frac{e^{x}-1-\sin x}{x^2}.

  1. −12-\dfrac{1}{2}
  2. 12\dfrac{1}{2} (correct answer)
  3. 11
  4. 00
  5. −1-1

Explanation: This limit is 0/0 indeterminate, so L'Hôpital's Rule applies. Differentiating gives (exe^x - cos x)/(2x), still 0/0; a second differentiation yields (exe^x + sin x)/2, evaluating to (1 + 0)/2 = 1/2. Taylor series confirms: e^x ≈ 1 + x + x²/2, sin x ≈ x, so numerator ≈ x²/2, divided by x² gives 1/2. A common mistake is to differentiate incorrectly, perhaps forgetting the chain rule for e^x. Another tempting error is assuming the limit is 0 by direct substitution without accounting for rates. A transferable strategy is to apply L'Hôpital's Rule until the form is determinate and cross-verify with series expansions for conceptual understanding.

Question 16

As x→0x\to 0, a damping factor is tan⁡x−xx3\frac{\tan x - x}{x^3}. What is the limit?

  1. 13\dfrac{1}{3} (correct answer)
  2. 00
  3. 11
  4. 12\dfrac{1}{2}
  5. 23\dfrac{2}{3}

Explanation: This limit has the indeterminate form 0/0 since both tan(x) - x and x³ approach 0 as x → 0. Applying L'Hôpital's Rule once gives (sec²(x) - 1)/(3x²), which can be rewritten as tan²(x)/(3x²) using the identity sec²(x) - 1 = tan²(x). This is still 0/0, so applying L'Hôpital's Rule again yields 2tan(x)sec²(x)/(6x). One more application gives 2sec²(x)(sec²(x) + 2tan²(x)sec²(x))/6, which evaluates to 2(1)(1 + 0)/6 = 1/3 as x → 0. A common error is not recognizing that sec²(x) - 1 = tan²(x), which simplifies the calculation. For functions like tan(x) - x near x = 0, expect the leading term in the Taylor expansion to be x³/3.

Question 17

Evaluate lim⁡x→0ex−1−xx2\displaystyle \lim_{x\to 0}\frac{e^{x}-1-x}{x^2}.

  1. 12\dfrac{1}{2} (correct answer)
  2. 11
  3. 00
  4. 22
  5. −12-\dfrac{1}{2}

Explanation: Lim (exe^x - 1 - x)/x^2 as x→0 is 0/0, L'Hôpital twice yields 1/2. A temptation is confusing with first-order approximation. The correct step uses e^x Taylor: x^2/2 term. This highlights second-order expansion. Verification matches 1/2. A transferable strategy is systematic Taylor use for exponentials.

Question 18

A decay model requires lim⁡x→0ln⁡(1+7x)−7xx2\lim_{x\to0}\frac{\ln(1+7x)-7x}{x^2}. What is the limit?

  1. −492-\frac{49}{2} (correct answer)
  2. 492\frac{49}{2}
  3. −49-49
  4. 00
  5. 77

Explanation: The expression ln(1 + 7x) - 7x approaches 0 as x approaches 0, creating a 0/0 indeterminate form. Applying L'Hôpital's Rule once gives (7/(1 + 7x) - 7)/(2x) = -49x/((1 + 7x)(2x)), which is still 0/0. Applying it again yields -49/(2(1 + 7x)) - 343x/(2(1 + 7x)²), which equals -49/2 when x = 0. A common error is forgetting that ln(1 + u) ≈ u - u²/2 for small u, which explains why the second-order term dominates. The strategy is to remember that when logarithms appear with their linear approximations subtracted, the limit often involves the coefficient squared divided by 2.

Question 19

Evaluate lim⁡x→0ex2−1x2\displaystyle \lim_{x\to 0}\frac{e^{x^2}-1}{x^2}.

  1. 22
  2. 11 (correct answer)
  3. 00
  4. 12\dfrac{1}{2}
  5. −1-1

Explanation: Indeterminate 0/0; L'Hôpital gives (2x ex2e^{x²})/(2x) = e^{x²} → 1. Equivalent to (eve^v -1)/v → 1, v=x². Tempting wrong: assuming 0. Another error: wrong derivative. Strategy: recognize standard limits in disguised forms.

Question 20

Find lim⁡x→01−cos⁡(9x)x2\displaystyle \lim_{x\to 0}\frac{1-\cos(9x)}{x^2}

  1. 8181
  2. 812\dfrac{81}{2} (correct answer)
  3. 99
  4. 00
  5. 162162

Explanation: Lim 1−cos⁡(9x)x2\frac{1 - \cos(9x)}{x^2} as x→0x \to 0 is 0/0, apply L'Hôpital twice. Derivatives lead to 812\frac{81}{2} via scaling the standard 12\frac{1}{2} limit by 929^2. A wrong move is forgetting the square, getting 9. The conceptual step uses 1−cos⁡(kx)x2=k22\frac{1 - \cos(kx)}{x^2} = \frac{k^2}{2}. This generalizes trigonometric limits. A transferable strategy is leveraging known limits with substitutions for efficiency.