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The World of Limits in Advanced Mathematics

L'Hôpital's Rule

L’Hôpital’s rule is an important method for solving 00\frac{0}{0} and ∞∞\frac{\infty}{\infty} indeterminate forms. It evaluates the limit of a ratio of functions by taking the limit of the ratio of their derivatives.

Basic Principle

L'Hôpital's rule

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 lim⁡x→af(x)=∞\lim_{x \to a} f(x) = \infty and lim⁡x→ag(x)=∞\lim_{x \to a} g(x) = \infty, 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)

Applicable Conditions

1. The 00\frac{0}{0} Indeterminate Form

  • The numerator vanishes at the limit point
  • The denominator vanishes at the limit point
  • Both numerator and denominator are differentiable

2. The ∞∞\frac{\infty}{\infty} Indeterminate Form

  • The numerator is infinite at the limit point
  • The denominator is infinite at the limit point
  • Both numerator and denominator are differentiable

3. Other Indeterminate Forms

  • 0⋅∞0 \cdot \infty: transform into 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}
  • ∞−∞\infty - \infty: transform into 00\frac{0}{0} or ∞∞\frac{\infty}{\infty} by combining fractions
  • 000^0, ∞0\infty^0, 1∞1^\infty: take logarithms to transform into 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}

Steps for Use

  1. Check the type of indeterminate form
  2. Differentiate the numerator and the denominator
  3. Evaluate the limit of the derivatives
  4. If it is still indeterminate, repeat steps 2–3

Worked Examples

Example 1

Find the limit lim⁡x→0sin⁡xx\lim_{x \to 0} \frac{\sin x}{x}

Reference Answer(1 个标签)
L'Hôpital's rule

Idea: This is a 00\frac{0}{0} indeterminate form, so we can use L’Hôpital’s rule.

Detailed steps:

  1. Check the type of indeterminate form:

    • As x→0x \to 0, the numerator sin⁡x→0\sin x \to 0
    • As x→0x \to 0, the denominator x→0x \to 0
    • Hence it is a 00\frac{0}{0} indeterminate form

  2. Differentiate the numerator and the denominator:

    • (sin⁡x)′=cos⁡x(\sin x)' = \cos x
    • (x)′=1(x)' = 1

  3. Evaluate the limit of the derivatives: lim⁡x→0cos⁡x1=cos⁡0=1\lim_{x \to 0} \frac{\cos x}{1} = \cos 0 = 1

Answer: lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1

Example 2

Find the limit lim⁡x→∞x2ex\lim_{x \to \infty} \frac{x^2}{e^x}

Reference Answer(1 个标签)
L'Hôpital's rule

Idea: This is a ∞∞\frac{\infty}{\infty} indeterminate form, so we can use L’Hôpital’s rule.

Detailed steps:

  1. Check the type of indeterminate form:

    • As x→∞x \to \infty, the numerator x2→∞x^2 \to \infty
    • As x→∞x \to \infty, the denominator ex→∞e^x \to \infty
    • Hence it is a ∞∞\frac{\infty}{\infty} indeterminate form

  2. Differentiate the numerator and the denominator:

    • (x2)′=2x(x^2)' = 2x
    • (ex)′=ex(e^x)' = e^x

  3. Evaluate the limit of the derivatives: lim⁡x→∞2xex\lim_{x \to \infty} \frac{2x}{e^x} This is still a ∞∞\frac{\infty}{\infty} indeterminate form

  4. Apply L’Hôpital’s rule again:

    • (2x)′=2(2x)' = 2
    • (ex)′=ex(e^x)' = e^x

  5. Evaluate the limit of the second derivatives: lim⁡x→∞2ex=0\lim_{x \to \infty} \frac{2}{e^x} = 0

Answer: lim⁡x→∞x2ex=0\lim_{x \to \infty} \frac{x^2}{e^x} = 0

Practice Problems

Exercise 1

Find the limit lim⁡x→0ex−1x\lim_{x \to 0} \frac{e^x - 1}{x}

Reference Answer(1 个标签)
L'Hôpital's rule

Idea: This is a 00\frac{0}{0} indeterminate form, so we can use L’Hôpital’s rule.

Detailed steps:

  1. Check the type of indeterminate form:

    • As x→0x \to 0, the numerator ex−1→0e^x - 1 \to 0
    • As x→0x \to 0, the denominator x→0x \to 0
    • Hence it is a 00\frac{0}{0} indeterminate form

  2. Differentiate the numerator and the denominator:

    • (ex−1)′=ex(e^x - 1)' = e^x
    • (x)′=1(x)' = 1

  3. Evaluate the limit of the derivatives: lim⁡x→0ex1=e0=1\lim_{x \to 0} \frac{e^x}{1} = e^0 = 1

Answer: lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1

Exercise 2

Find the limit lim⁡x→0ln⁡(1+x)x\lim_{x \to 0} \frac{\ln(1 + x)}{x}

Reference Answer(1 个标签)
L'Hôpital's rule

Idea: This is a 00\frac{0}{0} indeterminate form, so we can use L’Hôpital’s rule.

Detailed steps:

  1. Check the type of indeterminate form:

    • As x→0x \to 0, the numerator ln⁡(1+x)→0\ln(1 + x) \to 0
    • As x→0x \to 0, the denominator x→0x \to 0
    • Hence it is a 00\frac{0}{0} indeterminate form

  2. Differentiate the numerator and the denominator:

    • (ln⁡(1+x))′=11+x(\ln(1 + x))' = \frac{1}{1 + x}
    • (x)′=1(x)' = 1

  3. Evaluate the limit of the derivatives: lim⁡x→011+x1=lim⁡x→011+x=1\lim_{x \to 0} \frac{\frac{1}{1 + x}}{1} = \lim_{x \to 0} \frac{1}{1 + x} = 1

Answer: lim⁡x→0ln⁡(1+x)x=1\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1

Summary

Symbols Used in This Article

符号类型读音/说明在本文中的含义
lim⁡x→a\lim_{x \to a}数学符号limit as x approaches aThe limit as x→ax \to a
f(x),g(x)f(x), g(x)数学符号f/g of xNumerator and denominator functions
f′(x),g′(x)f'(x), g'(x)数学符号f/g prime of xDerivatives of the numerator and denominator
00\frac{0}{0}数学符号zero over zeroIndeterminate form
∞∞\frac{\infty}{\infty}数学符号infinity over infinityIndeterminate form
0⋅∞0 \cdot \infty数学符号zero times infinityIndeterminate form
∞−∞\infty - \infty数学符号infinity minus infinityIndeterminate form
00,∞0,1∞0^0, \infty^0, 1^\infty数学符号indeterminate formsPower-type indeterminate forms
sin⁡x,cos⁡x\sin x, \cos x数学符号sine/cosine of xTrigonometric functions
exe^x数学符号e to the xThe exponential function
ln⁡(1+x)\ln(1 + x)数学符号natural log of 1 plus xThe natural logarithm function

中英对照

中文术语英文术语音标说明
洛必达法则L’Hôpital’s rule/ˌləʊpiːˈtɑːlz ruːl/A method of evaluating limits via the ratio of derivatives
不定式indeterminate form/ˌɪndɪˈtɜːmɪnət fɔːm/A limit form that cannot be determined directly
导数derivative/dɪˈrɪvətɪv/The rate of change of a function at a point
可导differentiable/ˌdɪfəˈrenʃiəbl/The property that a derivative exists
无穷大infinity/ɪnˈfɪnəti/An unbounded quantity
通分common denominator/ˈkɒmən dɪˈnɒmɪneɪtə/The process of converting to a common denominator
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