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Integration Techniques, L’Hopital’s Rule, and Improper Integrals

8.7 Indeterminate Forms and L’Hopital’s Rule

Focus on indeterminate products, variable-power limits, l’hopital’s rule, other indeterminate forms, repeated applications.

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Ideas in this lesson

01Indeterminate productsWorked steps · five questions02Variable-power limitsWorked steps · five questions03L’Hopital’s RuleWorked steps · five questions04Other indeterminate formsWorked steps · five questions05Repeated applicationsWorked steps · five questions
Idea 1 of 5

Indeterminate products

Rewrite xln⁡xx \ln x as ln⁡x\ln x divided by 1x\frac{1}{x} before applying LL’Hopital.

Worked example 1

Evaluate lim⁡x→0+2xln⁡x\displaystyle \lim\limits_{x\to 0^{+}} 2x \ln x.
  1. xln⁡x=ln⁡x1xx \ln x = \dfrac{\ln x}{\dfrac{1}{x}}; form −∞∞\dfrac{-\infty }{\infty }
  2. lim⁡ln⁡x1x=lim⁡1x−1x2=lim⁡(−x)=0\displaystyle \lim \dfrac{\ln x}{\dfrac{1}{x}} = \lim \dfrac{\dfrac{1}{x}}{-\dfrac{1}{x^{2}}} = \lim (-x) = 0
  3. lim⁡x→0+2xln⁡x=2⋅0=0\displaystyle \lim\limits_{x\to 0^{+}} 2x \ln x = 2 \cdot 0 = 0

Answer: 00

Worked example 2

Evaluate lim⁡x→0+3xln⁡x\displaystyle \lim\limits_{x\to 0^{+}} 3x \ln x.
  1. xln⁡x=ln⁡x1xx \ln x = \dfrac{\ln x}{\dfrac{1}{x}}; form −∞∞\dfrac{-\infty }{\infty }
  2. lim⁡ln⁡x1x=lim⁡1x−1x2=lim⁡(−x)=0\displaystyle \lim \dfrac{\ln x}{\dfrac{1}{x}} = \lim \dfrac{\dfrac{1}{x}}{-\dfrac{1}{x^{2}}} = \lim (-x) = 0
  3. lim⁡x→0+3xln⁡x=3⋅0=0\displaystyle \lim\limits_{x\to 0^{+}} 3x \ln x = 3 \cdot 0 = 0

Answer: 00

Idea 2 of 5

Variable-power limits

Take logarithms of a positive variable power. Evaluate the logarithmic limit, then exponentiate.

Worked example 1

Evaluate lim⁡x→0(1+2x)1x\displaystyle \lim\limits_{x\to 0} (1 + 2x)^{\dfrac{1}{x}}.
  1. y=(1+2x)1xy = (1 + 2x)^{\dfrac{1}{x}}; ln⁡y=ln⁡(1+2x)x\ln y = \dfrac{\ln (1 + 2x)}{x}
  2. lim⁡ln⁡y=lim⁡21+2x=2\displaystyle \lim \ln y = \lim \dfrac{2}{1 + 2x} = 2
  3. lim⁡y=e2=e2≈7.39\displaystyle \lim y = e^{2} = e^{2} \approx 7.39

Answer: e2≈7.39e^{2} \approx 7.39

Worked example 2

Evaluate lim⁡x→0(1+3x)1x\displaystyle \lim\limits_{x\to 0} (1 + 3x)^{\dfrac{1}{x}}.
  1. y=(1+3x)1xy = (1 + 3x)^{\dfrac{1}{x}}; ln⁡y=ln⁡(1+3x)x\ln y = \dfrac{\ln (1 + 3x)}{x}
  2. lim⁡ln⁡y=lim⁡31+3x=3\displaystyle \lim \ln y = \lim \dfrac{3}{1 + 3x} = 3
  3. lim⁡y=e3=e3≈20.09\displaystyle \lim y = e^{3} = e^{3} \approx 20.09

Answer: e3≈20.09e^{3} \approx 20.09

Idea 3 of 5

L’Hopital’s Rule

Apply the rule only after verifying a00a \frac{0}{0} or ∞∞\frac{\infty }{\infty } form and its hypotheses. Differentiate numerator and denominator separately.

Worked example 1

Evaluate lim⁡x→0sin⁡(2x)x\displaystyle \lim\limits_{x\to 0} \dfrac{\sin (2x)}{x} using LL’Hopital’ss Rule.
  1. Numerator →0\to 0; denominator →0  ⟹  \to 0 \implies form 00\dfrac{0}{0}
  2. lim⁡x→02cos⁡(2x)1\displaystyle \lim\limits_{x\to 0} \dfrac{2 \cos (2x)}{1}
  3. 2cos⁡0=22 \cos 0 = 2

Answer: 22

Worked example 2

Evaluate lim⁡x→0e3x−1x\displaystyle \lim\limits_{x\to 0} \dfrac{e^{3x} - 1}{x} using LL’Hopital’ss Rule.
  1. Numerator →0\to 0; denominator →0  ⟹  \to 0 \implies form 00\dfrac{0}{0}
  2. lim⁡x→03e3x1\displaystyle \lim\limits_{x\to 0} \dfrac{3e^{3x}}{1}
  3. 3e0=33e^{0} = 3

Answer: 33

Idea 4 of 5

Other indeterminate forms

Rewrite products, differences, or variable powers as a quotient or logarithmic limit before applying the rule.

Worked example 1

Is ∞∞\dfrac{\infty }{\infty } an indeterminate form?
  1. lim⁡xx=1,lim⁡x2x=∞,lim⁡xx2=0(x→∞)\displaystyle \dfrac{\lim x}{x} = 1, \dfrac{\lim x^{2}}{x} = \infty , \dfrac{\lim x}{x^{2}} = 0 (x \to \infty )
  2. ∞∞  ⟹  \dfrac{\infty }{\infty } \implies indeterminate: distinct example limits above

Answer: Yes

Worked example 2

Is 0⋅∞0 \cdot \infty an indeterminate form?
  1. lim⁡x(1x)=1,lim⁡x(1x)=0(x→0+)\displaystyle \lim x(\dfrac{1}{x}) = 1, \lim x(\dfrac{1}{\sqrt{x}}) = 0 (x \to 0^{+})
  2. 0⋅∞  ⟹  0 \cdot \infty \implies indeterminate: distinct example limits above

Answer: Yes

Idea 5 of 5

Repeated applications

A first application can leave another indeterminate form. Recheck the form before applying the rule again.

Worked example 1

Evaluate lim⁡x→0sin⁡(2x)x\displaystyle \lim\limits_{x\to 0} \dfrac{\sin (2x)}{x} using LL’Hopital’ss Rule.
  1. Numerator →0\to 0; denominator →0  ⟹  \to 0 \implies form 00\dfrac{0}{0}
  2. lim⁡x→02cos⁡(2x)1\displaystyle \lim\limits_{x\to 0} \dfrac{2 \cos (2x)}{1}
  3. 2cos⁡0=22 \cos 0 = 2

Answer: 22

Worked example 2

Evaluate lim⁡x→0e3x−1x\displaystyle \lim\limits_{x\to 0} \dfrac{e^{3x} - 1}{x} using LL’Hopital’ss Rule.
  1. Numerator →0\to 0; denominator →0  ⟹  \to 0 \implies form 00\dfrac{0}{0}
  2. lim⁡x→03e3x1\displaystyle \lim\limits_{x\to 0} \dfrac{3e^{3x}}{1}
  3. 3e0=33e^{0} = 3

Answer: 33

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §8.7,p8.7, p. 557557. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.