L'Hospital's Rule - AP Calculus BC

Evaluate $\displaystyle\lim_{x\to 0}\dfrac{\arctan(3x)}{x}$ for a small-angle inverse tangent model.
Answer: $3$ 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.
A rate comparison is modeled by $\frac{\sqrt{x+9}-3}{x}$ as $x\to 0$ . Evaluate the limit.
Answer: $\frac{1}{6}$ This problem requires the use of L'Hôpital's Rule to evaluate the limit. As $x$ approaches 0, $\sqrt{x+9} - 3$ approaches 0 and $x$ approaches 0, yielding the $\frac{0}{0}$ indeterminate form, justifying L'Hôpital's Rule. Differentiating gives $\frac{1}{2\sqrt{x+9}}$ , which at $x=0$ is $\frac{1}{2 \times 3} = \frac{1}{6}$ . Rationalizing also confirms this result. A tempting distractor is 0, possibly from direct substitution without handling the form. To recognize indeterminate forms transferably, always substitute the limiting value into the numerator and denominator separately to check for $\frac{0}{0}$ or $\infty/\infty$ .
Evaluate $\displaystyle \lim_{x\to 0}\frac{x-\sin x}{x^3}$ appearing in an error estimate for small $x$ .
Answer: $\dfrac{1}{6}$ 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.
A damping ratio is modeled by $\dfrac{1-\cos(6x)}{x^2}$ ; find $\displaystyle \lim_{x\to 0}$ of this expression.
Answer: $18$ 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.
As $t \to 0$ , a model uses $\dfrac{e^{5t}-1}{t}$ ; what is the limit?
Answer: $5$ 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.
Compute $\displaystyle\lim_{x\to 0}\dfrac{e^{x^2}-1}{x^2}$ for a quadratic exponent perturbation.
Answer: $1$ This limit problem requires L'Hôpital's Rule to evaluate an indeterminate form. As x approaches 0, $e^{x^2}$ - 1 approaches 1-1=0 and $x^2$ approaches 0, forming 0/0. Derivatives: $e^{x^2}$ * 2x over 2x = $e^{x^2}$ , limit $e^0$ =1. The rule is appropriate as derivatives simplify to a direct limit. A tempting distractor is 0, from substituting without differentiation. To recognize indeterminate forms like 0/0, always plug in the limit value to check if both numerator and denominator approach 0 or infinity.
Find $\displaystyle\lim_{x\to 0}\dfrac{\ln(1+\sin x)}{x}$ for a composite small-change model.
Answer: $1$ This limit problem requires L'Hôpital's Rule to evaluate an indeterminate form. As x approaches 0, ln(1+sin x) approaches 0 and x approaches 0, 0/0. Derivatives: (1/(1+sin x)) * cos x / 1, limit 1*1=1. The form allows the rule. 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.
A material's strain ratio is $\frac{1-\cos(6x)}{x^2}$ near $x=0$ . Find $\lim_{x\to 0}\frac{1-\cos(6x)}{x^2}$ .
Answer: $18$ This problem requires the use of L'Hôpital's Rule to evaluate the limit. As $x$ approaches 0, $1 - \cos(6x)$ approaches 0 and $x^2$ approaches 0, forming the indeterminate $\frac{0}{0}$ , allowing L'Hôpital's Rule. First application gives $\frac{6 \sin(6x)}{2x}$ , which is still $\frac{0}{0}$ , so apply again to get $\frac{36 \cos(6x)}{2} = 18 \cos(6x)$ , evaluating to 18 at $x=0$ . This uses the rule multiple times until the form is determinate. A tempting distractor is 36, which might occur from forgetting to divide by 2 after the second differentiation. To recognize indeterminate forms transferably, always substitute the limiting value into the numerator and denominator separately to check for $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .
Compute $\displaystyle\lim_{x\to 0}\dfrac{\arcsin(2x)}{x}$ for a small-input inverse sine response.
Answer: $2$ This limit problem requires L'Hôpital's Rule to evaluate an indeterminate form. As x approaches 0, arcsin(2x) approaches 0 and x approaches 0, creating 0/0. Derivatives: 2 / √(1- $(2x)^2$ ) over 1, limit 2/1=2. The rule is justified by the indeterminate nature. A tempting distractor is 1, ignoring the coefficient in the argument. To recognize indeterminate forms like 0/0, always plug in the limit value to check if both numerator and denominator approach 0 or infinity.
Compute $\displaystyle\lim_{x\to 0}\dfrac{(1+x)^5-1}{x}$ , a limit from a binomial growth model.
Answer: $5$ The limit of ( $(1+x)^5$ - 1)/x as x approaches 0 uses L'Hôpital's Rule. 0/0, derivatives 5 $(1+x)^4$ / 1 = 5. Indeterminate justifies. One step. The distractor 1 ignores the power. Spot indeterminate by evaluation.
A thermodynamics simplification uses $\displaystyle \lim_{x\to 0}\frac{\arctan(2x)-2x}{x^3}$ . What is the limit?
Answer: $-\dfrac{8}{3}$ This problem requires L'Hôpital's Rule for the 0/0 indeterminate form. When x = 0, arctan(0) - 0 = 0 and 0³ = 0, creating the indeterminate form. Applying L'Hôpital's Rule once gives (2/(1+4x²) - 2)/(3x²), which simplifies to (2 - 2(1+4x²))/(3x²(1+4x²)) = -8x²/(3x²(1+4x²)) = -8/(3(1+4x²)). At x = 0, this equals -8/3. The answer 0 might mislead students who think arctan(2x) and 2x cancel exactly, but arctan has a negative cubic term in its Taylor expansion. For inverse trigonometric functions minus their linear approximations, the third-order term typically dominates near zero.
A population's relative change is $\frac{e^{3x}-1}{x}$ for small $x$ . Find $\lim_{x\to 0}\frac{e^{3x}-1}{x}$ .
Answer: $3$ This problem requires the use of L'Hôpital's Rule to evaluate the limit. As x approaches 0, $e^{3x} - 1$ approaches 0 and x approaches 0, yielding the indeterminate form $0/0$ , justifying the application of L'Hôpital's Rule. Differentiating gives $3 e^{3x}$ in the numerator and 1 in the denominator, so the limit is $3 e^{0} = 3 * 1 = 3$ . This matches the derivative of $e^{3x}$ at x=0, which is the instantaneous rate of change. A tempting distractor is 1, which could come from forgetting the chain rule and treating it as $(e^x - 1)/x$ without the 3. To recognize indeterminate forms transferably, always substitute the limiting value into the numerator and denominator separately to check for $0/0$ or $\infty/\infty$ .
Find $\displaystyle\lim_{x\to 0}\dfrac{e^{2x}-1}{\sin(5x)}$ for a rate ratio near $x=0$ .
Answer: $\dfrac{2}{5}$ L'Hôpital's Rule computes the limit of $(e^{2x} - 1)/\sin(5x)$ as x approaches 0. $0/0$ , derivatives $2 e^{2x} / 5 \cos(5x) = 2/5$ . Form allows the rule. Direct resolution. The distractor $5/2$ inverts the ratio mistakenly. Identify indeterminate forms via substitution check.
Evaluate $\displaystyle\lim_{x\to 0}\dfrac{\sqrt{9+x}-3}{x}$ for the instantaneous rate of change of a root function.
Answer: $\dfrac{1}{6}$ 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.
Compute $\displaystyle\lim_{x\to 0}\dfrac{e^{x}-1-x}{x^2}$ , appearing in a Taylor remainder bound.
Answer: $\dfrac{1}{2}$ L'Hôpital's Rule applies to the limit of ( $e^x$ - 1 - x)/ $x^2$ as x approaches 0. It's 0/0, differentiating to $e^x$ - 1 / (2x), still 0/0. Second derivatives give $e^x$ / 2 = 1/2. The rule is justified by the ongoing indeterminate form. The distractor 0 might result from ignoring higher-order terms. Always substitute to confirm 0/0 before applying the rule.
Find $\displaystyle\lim_{x\to 0}\dfrac{\sqrt{1+4x}-\sqrt{1+x}}{x}$ for a difference quotient of radicals.
Answer: $\dfrac{3}{2}$ This limit problem requires L'Hôpital's Rule to evaluate an indeterminate form. As x approaches 0, $\sqrt{1+4x} - \sqrt{1+x}$ approaches $1-1=0$ and x approaches 0, yielding $\frac{0}{0}$ . Derivatives: $\frac{4}{2\sqrt{1+4x}} - \frac{1}{2\sqrt{1+x}}$ over 1, at 0: $(4/2 - 1/2) = 2 - 0.5 = 1.5 = \frac{3}{2}$ . The rule directly gives the limit after differentiation. A tempting distractor is $\frac{1}{2}$ , from averaging coefficients incorrectly. To recognize indeterminate forms like $\frac{0}{0}$ , always plug in the limit value to check if both numerator and denominator approach 0 or infinity.
Compute $\displaystyle\lim_{x\to 0}\dfrac{\sin(7x)}{\sin(2x)}$ for comparing two oscillations near equilibrium.
Answer: $\dfrac{7}{2}$ The limit of sin(7x)/sin(2x) as x approaches 0 requires L'Hôpital's Rule. It's 0/0, derivatives 7 cos(7x)/ (2 cos(2x)) = 7/2 at 0. The indeterminate form allows this direct application. No further steps needed. The distractor 1 tempts if coefficients are ignored. Confirm indeterminate forms by substituting the approach value.
Evaluate $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{\sqrt{1+x}-1}$ , a ratio comparing two small changes.
Answer: $2$ 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.
Evaluate $\displaystyle\lim_{x\to 0}\dfrac{\tan(2x)}{x}$ for a small-angle approximation in radians.
Answer: $2$ 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.
Evaluate $\displaystyle\lim_{x\to 0}\dfrac{1-\cos x}{x\sin x}$ , a limit from a geometry-based derivation.
Answer: $\dfrac{1}{2}$ L'Hôpital's Rule evaluates the limit of (1 - cos x)/(x sin x) as x approaches 0. It's 0/0, derivatives sin x / (sin x + x cos x) = 1/2 at limit after simplification or further check. The form justifies application. Resolved effectively. The distractor 0 fails if higher terms ignored. Recognize indeterminate forms by evaluating components separately.

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