AP Calculus BC Quiz: Lhospitals Rule

20 questions · exam conditions

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Lhospitals RuleQuestion 1 of 20

Compute $\displaystyle \lim_{x\to \infty}\frac{\ln x}{\sqrt{x}}$ describing a slow-growth comparison.

\infty

1

0

\dfrac{1}{2}

\ln(\infty)

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AP Calculus BC Quiz

AP Calculus BC Quiz: Lhospitals Rule

Practice Lhospitals Rule in AP Calculus BC 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 BC.

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

Compute $\displaystyle \lim_{x\to \infty}\frac{\ln x}{\sqrt{x}}$ describing a slow-growth comparison.

$\infty$
$1$
$0$ (correct answer)
$\dfrac{1}{2}$
$\ln(\infty)$

Explanation: This problem applies L'Hôpital's Rule to the limit as x approaches infinity of ln(x)/√x. As x goes to infinity, it's ∞/∞, allowing the rule. Derivatives are (1/x)/(1/(2√x)) = 2/√x, which approaches 0. Logarithmic growth is slower than any positive power root. A tempting distractor is ∞, mistakenly thinking ln x grows faster than √x. To recognize indeterminate forms at infinity, assess if both numerator and denominator tend to infinity or zero in a way that creates ambiguity.

Question 2

Compute $\displaystyle \lim_{x\to 0}\frac{\sqrt{1+9x}-1}{x}$ for a linearized measurement formula.

$0$
$\dfrac{9}{2}$ (correct answer)
$9$
$\dfrac{1}{9}$
$\dfrac{1}{2}$

Explanation: This problem applies L'Hôpital's Rule to the limit as x approaches 0 of $\frac{\sqrt{1+9x} - 1}{x}$ . It results in $\frac{1-1}{0} = \frac{0}{0}$ , justifying the rule. Derivative is $\frac{9}{2\sqrt{1+9x}} / 1$ , evaluating to $\frac{9}{2}$ at x=0. Rationalizing by multiplying conjugate also yields $\frac{9}{2}$ . A tempting distractor is $9$ , from forgetting the 1/2 in the square root derivative. To identify $0/0$ forms, plug in the limit value and ensure it's indeterminate before differentiating.

Question 3

Evaluate $\displaystyle \lim_{x\to 0}\frac{x-\sin x}{x^3}$ appearing in an error estimate for small $x$ .

$0$
$\dfrac{1}{6}$ (correct answer)
$-\dfrac{1}{6}$
$1$
$-1$

Explanation: This problem uses L'Hôpital's Rule to find the limit as x approaches 0 of (x - sin x)/x^3. Substituting gives 0/0, an indeterminate form for the rule. First derivatives (1 - cos x)/( $3x^2$ ) are 0/0; second (sin x)/(6x) are 0/0; third (cos x)/6 = 1/6 at x=0. This aligns with sin x ≈ x - x^3/6 from Taylor series. A tempting distractor is -1/6, possibly from sign error in series. Recognize persistent indeterminate forms by applying the rule multiple times until resolved.

Question 4

Evaluate $\displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^3}$ for a higher-order small-angle correction.

$0$
$\dfrac{1}{3}$ (correct answer)
$-\dfrac{1}{3}$
$1$
$\dfrac{1}{6}$

Explanation: L'Hôpital's Rule evaluates the limit of $\frac{\tan x - x}{x^3}$ as x approaches 0. It's $\frac{0}{0}$ , first derivatives $\sec^2 x - \frac{1}{3x^2}$ , still $\frac{0}{0}$ . Second: $\frac{2 \sec^2 x \tan x}{6x}$ , $\frac{0}{0}$ . Third: complex but evaluates to $\frac{1}{3}$ . Repeated indeterminates justify the process. The distractor $0$ might come from premature stopping. Spot indeterminate forms by checking limits of numerator and denominator independently.

Question 5

Compute $\displaystyle\lim_{x\to 0}\dfrac{\sin(2x)-2\sin x}{x^3}$ for a third-order comparison near zero.

$0$
$1$
$-1$ (correct answer)
$2$
$-2$

Explanation: This limit problem requires L'Hôpital's Rule to evaluate an indeterminate form. As x approaches 0, sin(2x) - 2 sin(x) approaches 0 - 0 = 0 and x^3 approaches 0, yielding 0/0. Derivatives: 2 cos(2x) - 2 cos(x) over 3x^2, still 0/0 (21 - 21=0); again: -4 sin(2x) + 2 sin(x) over 6x, still 0/0; again: -8 cos(2x) + 2 cos(x) over 6 = (-81 + 21)/6 = -6/6 = -1. Repeated application is needed due to persistent indeterminacy. A tempting distractor is 0, from direct substitution without accounting for higher derivatives. To recognize indeterminate forms like 0/0, always plug in the limit value to check if both numerator and denominator approach 0 or infinity.

Question 6

Find $\displaystyle\lim_{x\to 0}\dfrac{\sin x - x}{x^3}$ , which arises in truncation error analysis.

$0$
$\dfrac{1}{6}$
$-\dfrac{1}{6}$ (correct answer)
$1$
$-1$

Explanation: L'Hôpital's Rule is essential for the limit of (sin x - x)/x^3 as x approaches 0. Substitution yields 0/0, leading to derivatives cos x - 1 over 3x^2, still 0/0. A second application gives -sin x / (6x), again 0/0, and a third gives -cos x / 6 = -1/6. The persistent indeterminate form justifies repeated use of the rule. The distractor 0 might arise from stopping after the first differentiation. To spot indeterminate forms, evaluate step-by-step after each rule application.

Question 7

Evaluate $\displaystyle\lim_{x\to 0}\dfrac{\tan(2x)}{x}$ for a small-angle approximation in radians.

$0$
$1$
$2$ (correct answer)
$\dfrac{1}{2}$
Does not exist

Explanation: L'Hôpital's Rule helps evaluate the limit of tan(2x)/x as x approaches 0. Substitution gives 0/0, so we differentiate to 2 sec^2(2x) over 1, which is 2 at x=0. The indeterminate form justifies applying the rule to find this finite limit. No additional applications are required here. The distractor 1 might tempt if one forgets the factor of 2 from the chain rule. A strategy for spotting indeterminate forms is to compute numerator and denominator separately at the limit value.

Question 8

Compute $\displaystyle\lim_{x\to 0}\dfrac{e^{4x}-1}{x}$ , an indeterminate form arising in a growth-rate calculation.

$0$
$4$ (correct answer)
$1$
$e^{4}$
$\infty$

Explanation: L'Hôpital's Rule is the key skill for computing the limit of $(e^{4x} - 1)/x$ as x approaches 0. Substituting x=0 produces $0/0$ , an indeterminate form that allows us to apply the rule by differentiating numerator and denominator. The derivative of the numerator is $4e^{4x}$ and of the denominator is 1, yielding $4e^{0} = 4$ . This direct application resolves the limit without further steps. The distractor $e^4$ might arise from incorrectly exponentiating instead of differentiating properly. Always check for $0/0$ or $\infty/\infty$ by plugging in the limit value to confirm if L'Hôpital's Rule applies.

Question 9

Evaluate $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{\sqrt{1+x}-1}$ , a ratio comparing two small changes.

$0$
$2$ (correct answer)
$\dfrac{1}{2}$
$1$
$-2$

Explanation: This limit problem requires L'Hôpital's Rule to evaluate an indeterminate form. As x approaches 0, both the numerator sin(x) and the denominator √(1+x) - 1 approach 0, creating a 0/0 indeterminate form that justifies differentiating the numerator and denominator. The derivative of the numerator is cos(x), and the derivative of the denominator is 1/(2√(1+x)), so the limit becomes lim_{x→0} cos(x) / (1/(2√(1+x))) = 1 / (1/2) = 2. Applying L'Hôpital's Rule resolves the indeterminacy because the derivatives exist and the limit of the ratio exists. A tempting distractor might be 1/2, which could arise from mistakenly rationalizing the denominator without applying L'Hôpital correctly. To recognize indeterminate forms like 0/0, always plug in the limit value to check if both numerator and denominator approach 0 or infinity.

Question 10

Compute $\displaystyle\lim_{x\to 0}\dfrac{\ln(1+x)}{\sqrt{1+x}-1}$ for comparing two linearizations.

$2$ (correct answer)
$\dfrac{1}{2}$
$1$
$0$
$-2$

Explanation: This limit problem requires L'Hôpital's Rule to evaluate an indeterminate form. As x approaches 0, ln(1+x) approaches 0 and √(1+x) - 1 approaches 0, resulting in a 0/0 form that allows us to apply the rule. Differentiating gives 1/(1+x) in the numerator and 1/(2√(1+x)) in the denominator, so the limit is lim_{x→0} [1/(1+x)] / [1/(2√(1+x))] = 1 / (1/2) = 2. The rule is justified here because the original limit is indeterminate, but the derivatives lead to a determinate value. One tempting distractor is 1/2, possibly from inverting the derivative ratio incorrectly. To recognize indeterminate forms like 0/0, always plug in the limit value to check if both numerator and denominator approach 0 or infinity.

Question 11

Find $\displaystyle\lim_{x\to 0}\dfrac{\sin^2 x}{x^2}$ for a normalized energy term near zero.

$0$
$1$ (correct answer)
$2$
$\dfrac{1}{2}$
Does not exist

Explanation: This limit problem requires L'Hôpital's Rule to evaluate an indeterminate form, though actually it's not needed here since it's (sin x / x)^2 →1^2=1. As x→0, sin^2 x →0, x^2→0, 0/0, but L'Hôpital: 2 sin x cos x / 2x = sin x cos x / x, still 0/0; again: (cos x cos x + sin x (-sin x)) /1 = cos^2 x - sin^2 x →1. But directly known as 1. The form justifies, but simpler ways exist. A tempting distractor is 0, from substitution. To recognize indeterminate forms like 0/0, always plug in the limit value to check if both numerator and denominator approach 0 or infinity.

Question 12

Evaluate $\displaystyle\lim_{x\to 0}\dfrac{\sqrt{9+x}-3}{x}$ for the instantaneous rate of change of a root function.

$\dfrac{1}{6}$ (correct answer)
$\dfrac{1}{3}$
$\dfrac{1}{9}$
$0$
$6$

Explanation: L'Hôpital's Rule handles the limit of $\frac{\sqrt{9+x} - 3}{x}$ as x approaches 0. 0/0, derivatives $\frac{1}{2\sqrt{9+x}} / 1 = \frac{1}{6}$ . Form justifies use. Direct. The distractor $\frac{1}{3}$ doubles incorrectly. Identify indeterminate forms through plug-in.

Question 13

Compute $\displaystyle \lim_{x\to \infty}\frac{3x^2-5x+1}{6x^2+2x-7}$ using an appropriate method.

$0$
$\dfrac{1}{2}$ (correct answer)
$\dfrac{1}{3}$
$2$
DNE

Explanation: This limit involves L'Hôpital's Rule for the ∞/∞ indeterminate form. As x approaches infinity, both numerator and denominator approach infinity. Applying L'Hôpital's Rule once gives (6x - 5)/(12x + 2), still ∞/∞. Applying again yields 6/12 = 1/2. Alternatively, divide numerator and denominator by x² to get (3 - 5/x + 1/x²)/(6 + 2/x - 7/x²), which approaches 3/6 = 1/2 as x → ∞. The answer 1/3 might tempt those who incorrectly simplify or make arithmetic errors. For rational functions where degrees match, the limit at infinity equals the ratio of leading coefficients.

Question 14

To compare rates, compute $\displaystyle\lim_{x\to\infty}\frac{x}{\ln x}$ using an appropriate indeterminate-form method.

$0$
$1$
$\infty$ (correct answer)
$e$
$-\infty$

Explanation: This rate comparison limit requires L'Hôpital's Rule for the indeterminate form ∞/∞. As x → ∞, both x and ln(x) approach infinity, giving ∞/∞. Applying L'Hôpital's Rule: lim[x→∞] 1/(1/x) = lim[x→∞] x = ∞. Students might incorrectly think the limit is 0 by confusing which function grows faster. The key insight is that x grows faster than ln(x), so x/ln(x) → ∞ while ln(x)/x → 0, demonstrating the importance of correctly identifying which function dominates.

Question 15

As $t\to 0$ , the average power factor is modeled by $\frac{\sin(5t)}{t}$ . What is the limit?

$0$
$5$ (correct answer)
$\frac{1}{5}$
$\sin(0)$
$\cos(0)$

Explanation: This problem requires the use of L'Hôpital's Rule to evaluate the limit. As t approaches 0, both sin(5t) and t approach 0, creating the indeterminate form 0/0, which allows us to apply L'Hôpital's Rule by differentiating the numerator and denominator. The derivative of the numerator is 5 cos(5t), and the derivative of the denominator is 1, so the limit becomes 5 cos(5t) as t approaches 0. This evaluates to 5 cos(0) = 5 * 1 = 5. A tempting distractor is 1/5, which might result from mistakenly inverting the coefficient in the standard sin(u)/u limit without proper adjustment. To recognize indeterminate forms transferably, always substitute the limiting value into the numerator and denominator separately to check for 0/0 or ∞/∞.

Question 16

As $t \to 0$ , a model uses $\dfrac{e^{5t}-1}{t}$ ; what is the limit?

$0$
$1$
$5$ (correct answer)
$e^{5}$
$\infty$

Explanation: This problem involves applying L'Hôpital's Rule to evaluate the limit as $t \to 0$ of $\dfrac{e^{5t} - 1}{t}$ . As $t$ approaches 0, the numerator $e^{5t} - 1$ approaches 0 and the denominator $t$ approaches 0, creating the indeterminate form $\frac{0}{0}$ that justifies using L'Hôpital's Rule. Differentiating the numerator gives $5e^{5t}$ and the denominator gives 1, so the limit simplifies to $5e^{5t}$ evaluated at $t=0$ , which is 5. This result aligns with the derivative of $e^{5t}$ at $t=0$ , providing a conceptual check. A tempting distractor is $e^5$ , which might arise from mistakenly evaluating the numerator at $t=5$ instead of applying the rule properly. To recognize indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ in limits, substitute the limiting value and verify if the expression is undefined.

Question 17

A damping ratio is modeled by $\dfrac{1-\cos(6x)}{x^2}$ ; find $\displaystyle \lim_{x\to 0}$ of this expression.

$0$
$6$
$18$ (correct answer)
$36$
$\dfrac{1}{36}$

Explanation: This problem applies L'Hôpital's Rule to evaluate the limit as x approaches 0 of $\frac{1 - \cos(6x)}{x^2}$ . Substituting x=0 gives $\frac{1-1}{0} = \frac{0}{0}$ , an indeterminate form that permits the rule. First derivatives are $\frac{6 \sin(6x)}{2x}$ , still $\frac{0}{0}$ , so apply again to get $\frac{36 \cos(6x)}{2} = 18 \cos(0) = 18$ . Multiple applications are needed due to the persistent indeterminate form. A tempting distractor is 36, which might come from forgetting to divide by 2 after the second differentiation. Recognize indeterminate forms by substituting the limit value and repeatedly checking after each differentiation if needed.

Question 18

Find $\displaystyle \lim_{x\to 0}\frac{\sin(7x)}{x}$ for a small-angle approximation in a sensor.

$0$
$1$
$7$ (correct answer)
$\sin 7$
$\dfrac{1}{7}$

Explanation: This problem requires L'Hôpital's Rule to find the limit as x approaches 0 of $\sin(7x)/x$ . Substituting x=0 yields $\sin(0)/0 = 0/0$ , an indeterminate form that allows L'Hôpital's Rule to be applied. The derivatives are $7 \cos(7x)$ for the numerator and $1$ for the denominator, so the limit is $7 \cos(0) = 7$ . This is equivalent to 7 times the standard limit of $\sin(u)/u$ as u approaches 0, where $u=7x$ . A tempting distractor is 1, which ignores the coefficient 7 and treats it as the basic sine limit. Always check for indeterminate forms by plugging in the limit value and seeing if it results in $0/0$ or $\infty/\infty$ before proceeding with differentiation.

Question 19

A rate model uses $\dfrac{e^{2x}}{x}$ as $x\to\infty$ ; evaluate $\displaystyle \lim_{x\to\infty}\frac{e^{2x}}{x}$

$0$
$2$
$1$
$\infty$ (correct answer)
$e^{2}$

Explanation: This problem involves L'Hôpital's Rule for the limit as x approaches infinity of $\frac{e^{2x}}{x}$ . It presents as $\infty/\infty$ , justifying application. Derivative is $\frac{2e^{2x}}{1}$ , still $\infty/\infty$ , and repeated applications show exponential dominance, yielding $\infty$ . Exponentials grow faster than polynomials. A tempting distractor is 0, from incorrectly assuming the denominator overtakes. Spot indeterminate $\infty/\infty$ forms by evaluating asymptotic behavior and confirming unbounded growth in both parts.

Question 20

Evaluate $\displaystyle\lim_{x\to 0}\dfrac{\arctan(3x)}{x}$ for a small-angle inverse tangent model.

$\dfrac{1}{3}$
$0$
$3$ (correct answer)
$\pi$
$1$

Explanation: This limit problem requires L'Hôpital's Rule to evaluate an indeterminate form. As x approaches 0, arctan(3x) approaches 0 and x approaches 0, forming 0/0. Derivatives: 3/(1+(3x)^2) over 1, limit 3/1=3. The indeterminate form justifies the rule's application. A tempting distractor is 1, assuming the coefficient doesn't affect the limit. To recognize indeterminate forms like 0/0, always plug in the limit value to check if both numerator and denominator approach 0 or infinity.