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Calculus - L’Hôpital’s rule (HL)

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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L'Hôpital's Rule is a technique used in calculus to evaluate limits that result in indeterminate forms, specifically 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}.

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The rule states that if lim⁡x→af(x)=0\lim_{x \to a} f(x) = 0 and lim⁡x→ag(x)=0\lim_{x \to a} g(x) = 0 (or both are infinite), then lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}, provided the limit on the right exists or is infinite.

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For the rule to be applicable, the functions f(x)f(x) and g(x)g(x) must be differentiable on an open interval containing aa (except possibly at aa itself), and g′(x)≠0g'(x) \neq 0 near aa.

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L'Hôpital's Rule can be applied repeatedly. If the first application results in another indeterminate form, you can take the second derivatives: lim⁡x→af′′(x)g′′(x)\lim_{x \to a} \frac{f''(x)}{g''(x)}, and so on.

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Important: L'Hôpital's Rule is not the same as the Quotient Rule. You differentiate the numerator and the denominator separately.

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Other indeterminate forms like 0×∞0 \times \infty, ∞−∞\infty - \infty, 000^0, 1∞1^{\infty}, and ∞0\infty^0 must be algebraically transformed into 00\frac{0}{0} or ∞∞\frac{\infty}{\infty} before the rule can be applied.

📐Formulae

lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}

Indeterminate Forms: 00,∞∞\text{Indeterminate Forms: } \frac{0}{0}, \frac{\infty}{\infty}

💡Examples

Problem 1:

Evaluate the limit: lim⁡x→0sin⁡(5x)x\lim_{x \to 0} \frac{\sin(5x)}{x}.

Solution:

  1. Check for indeterminate form: as x→0x \to 0, sin⁡(5(0))=0\sin(5(0)) = 0 and the denominator is 00. This is the 00\frac{0}{0} form.
  2. Apply L'Hôpital's Rule: Differentiate the numerator and denominator.
  3. Let f(x)=sin⁡(5x)  ⟹  f′(x)=5cos⁡(5x)f(x) = \sin(5x) \implies f'(x) = 5\cos(5x).
  4. Let g(x)=x  ⟹  g′(x)=1g(x) = x \implies g'(x) = 1.
  5. lim⁡x→05cos⁡(5x)1=5cos⁡(0)=5(1)=5\lim_{x \to 0} \frac{5\cos(5x)}{1} = 5\cos(0) = 5(1) = 5.

Explanation:

Since the initial substitution resulted in 00\frac{0}{0}, we applied the rule by differentiating the top and bottom separately and then re-evaluating the limit.

Problem 2:

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

Solution:

  1. Substitute x=0x=0: e0−1−002=1−1−00=00\frac{e^0 - 1 - 0}{0^2} = \frac{1 - 1 - 0}{0} = \frac{0}{0}.
  2. First application of L'Hôpital's Rule: lim⁡x→0ddx(ex−1−x)ddx(x2)=lim⁡x→0ex−12x\lim_{x \to 0} \frac{\frac{d}{dx}(e^x - 1 - x)}{\frac{d}{dx}(x^2)} = \lim_{x \to 0} \frac{e^x - 1}{2x}
  3. Substitute x=0x=0 again: e0−12(0)=00\frac{e^0 - 1}{2(0)} = \frac{0}{0}. Still indeterminate.
  4. Second application of L'Hôpital's Rule: lim⁡x→0ddx(ex−1)ddx(2x)=lim⁡x→0ex2\lim_{x \to 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(2x)} = \lim_{x \to 0} \frac{e^x}{2}
  5. Evaluate: e02=12\frac{e^0}{2} = \frac{1}{2}.

Explanation:

This example demonstrates that L'Hôpital's Rule can be applied multiple times in succession until a determinate value is reached.

Problem 3:

Evaluate lim⁡x→∞ln⁡(x)x\lim_{x \to \infty} \frac{\ln(x)}{x}.

Solution:

  1. Check for indeterminate form: as x→∞x \to \infty, ln⁡(x)→∞\ln(x) \to \infty and x→∞x \to \infty. This is the ∞∞\frac{\infty}{\infty} form.
  2. Apply L'Hôpital's Rule: lim⁡x→∞1x1\lim_{x \to \infty} \frac{\frac{1}{x}}{1}
  3. Simplify: lim⁡x→∞1x=0\lim_{x \to \infty} \frac{1}{x} = 0.

Explanation:

Even though the limit is approaching infinity, the ratio of the rates of growth determines the limit. Since xx grows faster than ln⁡(x)\ln(x), the limit is 00.