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AP Calculus BC Flashcards: Lhospitals Rule

Study Lhospitals Rule in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Lhospitals Rule, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

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Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Calculus BC Flashcards: Lhospitals Rule

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QUESTION

Determine the form of lim⁡x→∞exx3\lim_{x \to \infty} \frac{e^x}{x^3}limx→∞​x3ex​ without evaluating.

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ANSWER

∞∞\frac{\infty}{\infty}∞∞​. Both ex→∞e^x \to \inftyex→∞ and x3→∞x^3 \to \inftyx3→∞ as x→∞x \to \inftyx→∞.

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Flashcard 1: Determine the form of lim⁡x→∞exx3\lim_{x \to \infty} \frac{e^x}{x^3}limx→∞​x3ex​ without evaluating.

Answer: ∞∞\frac{\infty}{\infty}∞∞​. Both ex→∞e^x \to \inftyex→∞ and x3→∞x^3 \to \inftyx3→∞ as x→∞x \to \inftyx→∞.

Flashcard 2: What must be true about f′(x)f'(x)f′(x) and g′(x)g'(x)g′(x) for L'Hospital's Rule to apply?

Answer: f′(x)f'(x)f′(x) and g′(x)g'(x)g′(x) must exist near ccc and g′(x)≠0g'(x) \neq 0g′(x)=0. Ensures the rule can be applied and gives a valid result.

Flashcard 3: Evaluate lim⁡x→0sin⁡2xx\lim_{{x \to 0}} \frac{\sin 2x}{x}limx→0​xsin2x​ using L'Hospital's Rule.

Answer:

  1. ddx(sin⁡2x)=2cos⁡2x\frac{d}{dx}(\sin 2x) = 2\cos 2xdxd​(sin2x)=2cos2x, so 2cos⁡01=2\frac{2\cos 0}{1} = 212cos0​=2.

Flashcard 4: Is L'Hospital's Rule applicable to lim⁡x→0x2sin⁡x\lim_{{x \to 0}} \frac{x^2}{\sin x}limx→0​sinxx2​?

Answer: Yes, the limit is in the form 00\frac{0}{0}00​. Both 02=00^2 = 002=0 and sin⁡0=0\sin 0 = 0sin0=0 at x=0x = 0x=0.

Flashcard 5: Evaluate lim⁡x→∞ln⁡xx\lim_{{x \to \infty}} \frac{\ln x}{x}limx→∞​xlnx​ using L'Hospital's Rule.

Answer:

  1. ddx(ln⁡x)=1x\frac{d}{dx}(\ln x) = \frac{1}{x}dxd​(lnx)=x1​ and ddx(x)=1\frac{d}{dx}(x) = 1dxd​(x)=1, so 1/x1→0\frac{1/x}{1} \to 011/x​→0.

Flashcard 6: Can L'Hospital's Rule be applied repeatedly?

Answer: Yes, if 00\frac{0}{0}00​ or ∞∞\frac{\infty}{\infty}∞∞​ persists after differentiation. Continue applying until a determinate form is reached.

Flashcard 7: Evaluate lim⁡x→∞2x2+3xx2−4\lim_{{x \to \infty}} \frac{2x^2 + 3x}{x^2 - 4}limx→∞​x2−42x2+3x​ using L'Hospital's Rule.

Answer:

  1. Apply L'Hospital's Rule: 4x+32x→42=2\frac{4x + 3}{2x} \to \frac{4}{2} = 22x4x+3​→24​=2.

Flashcard 8: Evaluate lim⁡x→0arcsin⁡xx\lim_{{x \to 0}} \frac{\arcsin x}{x}limx→0​xarcsinx​ using L'Hospital's Rule.

Answer:

  1. ddx(arcsin⁡x)=11−x2\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}dxd​(arcsinx)=1−x2​1​, so 11=1\frac{1}{1} = 111​=1.

Flashcard 9: Evaluate lim⁡x→0ln⁡(1+x)x\lim_{{x \to 0}} \frac{\ln(1+x)}{x}limx→0​xln(1+x)​ using L'Hospital's Rule.

Answer:

  1. ddx(ln⁡(1+x))=11+x\frac{d}{dx}(\ln(1+x)) = \frac{1}{1+x}dxd​(ln(1+x))=1+x1​, so 11=1\frac{1}{1} = 111​=1.

Flashcard 10: Evaluate lim⁡x→∞xx2+1\lim_{{x \to \infty}} \frac{x}{x^2 + 1}limx→∞​x2+1x​ using L'Hospital's Rule.

Answer:

  1. ddx(x)=1\frac{d}{dx}(x) = 1dxd​(x)=1 and ddx(x2+1)=2x\frac{d}{dx}(x^2 + 1) = 2xdxd​(x2+1)=2x, so 12x→0\frac{1}{2x} \to 02x1​→0.

Flashcard 11: State L'Hospital's Rule for limits of indeterminate forms.

Answer: lim⁡x→cf(x)g(x)=lim⁡x→cf′(x)g′(x)\lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)}limx→c​g(x)f(x)​=limx→c​g′(x)f′(x)​, if the limit exists. Take derivatives of numerator and denominator separately.

Flashcard 12: Evaluate lim⁡x→∞ln⁡xx2\lim_{{x \to \infty}} \frac{\ln x}{x^2}limx→∞​x2lnx​ using L'Hospital's Rule.

Answer:

  1. Apply L'Hospital's Rule: 1/x2x=12x2→0\frac{1/x}{2x} = \frac{1}{2x^2} \to 02x1/x​=2x21​→0.

Flashcard 13: Determine if L'Hospital's Rule applies: lim⁡x→∞x3ex\lim_{{x \to \infty}} \frac{x^3}{e^x}limx→∞​exx3​.

Answer: Yes, the limit is in the form ∞∞\frac{\infty}{\infty}∞∞​. Both x3→∞x^3 \to \inftyx3→∞ and ex→∞e^x \to \inftyex→∞ as x→∞x \to \inftyx→∞.

Flashcard 14: Find lim⁡x→0tan⁡xx\lim_{{x \to 0}} \frac{\tan x}{x}limx→0​xtanx​ using L'Hospital's Rule.

Answer:

  1. ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x) = \sec^2 xdxd​(tanx)=sec2x and ddx(x)=1\frac{d}{dx}(x) = 1dxd​(x)=1, so 11=1\frac{1}{1} = 111​=1.

Flashcard 15: What is the basic condition to apply L'Hospital's Rule?

Answer: The limit must be in the form 00\frac{0}{0}00​ or ∞∞\frac{\infty}{\infty}∞∞​. These are the only indeterminate forms where L'Hospital's Rule applies.

Flashcard 16: Find lim⁡x→∞x2ex\lim_{{x \to \infty}} \frac{x^2}{e^x}limx→∞​exx2​ using L'Hospital's Rule.

Answer:

  1. Apply L'Hospital's Rule twice to get 2ex→0\frac{2}{e^x} \to 0ex2​→0.

Flashcard 17: Determine if L'Hospital's Rule applies: lim⁡x→1x2−1x−1\lim_{{x \to 1}} \frac{x^2 - 1}{x - 1}limx→1​x−1x2−1​.

Answer: Yes, the limit is in the form 00\frac{0}{0}00​. Both (1)2−1=0(1)^2 - 1 = 0(1)2−1=0 and 1−1=01 - 1 = 01−1=0 at x=1x = 1x=1.

Flashcard 18: Find lim⁡x→0ex−1−xx2\lim_{{x \to 0}} \frac{e^x - 1 - x}{x^2}limx→0​x2ex−1−x​ using L'Hospital's Rule.

Answer: 12\frac{1}{2}21​. Apply L'Hospital's Rule twice: ex−12x→ex2=12\frac{e^x - 1}{2x} \to \frac{e^x}{2} = \frac{1}{2}2xex−1​→2ex​=21​.

Flashcard 19: Determine the form of lim⁡x→0x3ex−1\lim_{{x \to 0}} \frac{x^3}{e^x - 1}limx→0​ex−1x3​ without evaluating.

Answer: 00\frac{0}{0}00​. Both 03=00^3 = 003=0 and e0−1=0e^0 - 1 = 0e0−1=0 at x=0x = 0x=0.

Flashcard 20: Determine the form for lim⁡x→∞ln⁡xx\lim_{{x \to \infty}} \frac{\ln x}{\sqrt{x}}limx→∞​x​lnx​ without evaluating.

Answer: ∞∞\frac{\infty}{\infty}∞∞​. Both ln⁡x→∞\ln x \to \inftylnx→∞ and x→∞\sqrt{x} \to \inftyx​→∞ as x→∞x \to \inftyx→∞.

Flashcard 21: Evaluate lim⁡x→∞xex\lim_{{x \to \infty}} \frac{x}{e^x}limx→∞​exx​ using L'Hospital's Rule.

Answer:

  1. Apply L'Hospital's Rule: 1ex→0\frac{1}{e^x} \to 0ex1​→0 as x→∞x \to \inftyx→∞.

Flashcard 22: Is L'Hospital's Rule applicable to lim⁡x→2x2−4x−2\lim_{{x \to 2}} \frac{x^2 - 4}{x - 2}limx→2​x−2x2−4​?

Answer: Yes, the limit is in the form 00\frac{0}{0}00​. Both 4−4=04 - 4 = 04−4=0 and 2−2=02 - 2 = 02−2=0 at x=2x = 2x=2.

Flashcard 23: What is the result of lim⁡x→0ex−1x\lim_{{x \to 0}} \frac{e^x - 1}{x}limx→0​xex−1​ using L'Hospital's Rule?

Answer:

  1. ddx(ex−1)=ex\frac{d}{dx}(e^x - 1) = e^xdxd​(ex−1)=ex and ddx(x)=1\frac{d}{dx}(x) = 1dxd​(x)=1, so e01=1\frac{e^0}{1} = 11e0​=1.

Flashcard 24: Determine if L'Hospital's Rule applies: lim⁡x→∞x2ex\lim_{{x \to \infty}} \frac{x^2}{e^x}limx→∞​exx2​.

Answer: Yes, the limit is in the form ∞∞\frac{\infty}{\infty}∞∞​. Both x2→∞x^2 \to \inftyx2→∞ and ex→∞e^x \to \inftyex→∞ as x→∞x \to \inftyx→∞.

Flashcard 25: Identify the form lim⁡x→0xsin⁡x\lim_{{x \to 0}} \frac{x}{\sin x}limx→0​sinxx​ without evaluating.

Answer: 00\frac{0}{0}00​. Both numerator and denominator approach 0 as x→0x \to 0x→0.

Flashcard 26: Identify if L'Hospital's Rule applies: lim⁡x→0sin⁡xx\lim_{{x \to 0}} \frac{\sin x}{x}limx→0​xsinx​.

Answer: Yes, the limit is in the form 00\frac{0}{0}00​. Both sin⁡0=0\sin 0 = 0sin0=0 and 0=00 = 00=0, giving 00\frac{0}{0}00​ form.

Flashcard 27: What is the result of lim⁡x→01−cos⁡xx2\lim_{{x \to 0}} \frac{1 - \cos x}{x^2}limx→0​x21−cosx​ using L'Hospital's Rule?

Answer: 12\frac{1}{2}21​. Apply L'Hospital's Rule twice: sin⁡x2x→cos⁡x2=12\frac{\sin x}{2x} \to \frac{\cos x}{2} = \frac{1}{2}2xsinx​→2cosx​=21​.

Flashcard 28: Determine if L'Hospital's Rule applies: lim⁡x→0x−sin⁡xx3\lim_{{x \to 0}} \frac{x - \sin x}{x^3}limx→0​x3x−sinx​.

Answer: Yes, the limit is in the form 00\frac{0}{0}00​. Both 0−sin⁡0=00 - \sin 0 = 00−sin0=0 and 03=00^3 = 003=0 at x=0x = 0x=0.

Flashcard 29: Is L'Hospital's Rule applicable to lim⁡x→2x2−4x−2\lim_{{x \to 2}} \frac{x^2 - 4}{x - 2}limx→2​x−2x2−4​?

Answer: Yes, the limit is in the form 00\frac{0}{0}00​. Both 4−4=04 - 4 = 04−4=0 and 2−2=02 - 2 = 02−2=0 at x=2x = 2x=2.

Flashcard 30: Determine the form of lim⁡x→0x3ex−1\lim_{{x \to 0}} \frac{x^3}{e^x - 1}limx→0​ex−1x3​ without evaluating.

Answer: 00\frac{0}{0}00​. Both 03=00^3 = 003=0 and e0−1=0e^0 - 1 = 0e0−1=0 at x=0x = 0x=0.