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The World of Limits in Advanced Mathematics

Equivalent Infinitesimal Substitution

The method of equivalent infinitesimal substitution is an important technique for evaluating limits. By replacing a complicated infinitesimal with an equivalent simpler one, we can greatly simplify the computation of limits.

Basic Principle

If α(x)\alpha(x) and β(x)\beta(x) are equivalent infinitesimals, i.e., α(x)∼β(x)\alpha(x) \sim \beta(x), then:

Basic principle of equivalent infinitesimal substitution

lim⁡x→af(x)α(x)=lim⁡x→af(x)β(x)\lim_{x \to a} \frac{f(x)}{\alpha(x)} = \lim_{x \to a} \frac{f(x)}{\beta(x)}

Definition of Equivalent Infinitesimals

Definition of equivalent infinitesimals

As x→ax \to a, if lim⁡x→aα(x)β(x)=1\lim_{x \to a} \frac{\alpha(x)}{\beta(x)} = 1, then α(x)\alpha(x) and β(x)\beta(x) are called equivalent infinitesimals, denoted α(x)∼β(x)\alpha(x) \sim \beta(x).

Common Equivalent Infinitesimal Relations

1. Basic Equivalences (as x→0x \to 0)

sin⁡x∼x\sin x \sim x tan⁡x∼x\tan x \sim x arcsin⁡x∼x\arcsin x \sim x arctan⁡x∼x\arctan x \sim x ln⁡(1+x)∼x\ln(1 + x) \sim x ex−1∼xe^x - 1 \sim x (1+x)a−1∼ax(1 + x)^a - 1 \sim ax

2. Composite Equivalences

sin⁡(x2)∼x2\sin(x^2) \sim x^2 ln⁡(1+x2)∼x2\ln(1 + x^2) \sim x^2 ex2−1∼x2e^{x^2} - 1 \sim x^2

3. Higher-Order Equivalences

sin⁡x−x∼−x36\sin x - x \sim -\frac{x^3}{6} tan⁡x−x∼x33\tan x - x \sim \frac{x^3}{3} ln⁡(1+x)−x∼−x22\ln(1 + x) - x \sim -\frac{x^2}{2}

Worked Examples

Example 1

Find the limit lim⁡x→0sin⁡3xtan⁡2x\lim_{x \to 0} \frac{\sin 3x}{\tan 2x}

equivalent infinitesimal

Approach: This is a 00\frac{0}{0} indeterminate form, so we can use equivalent infinitesimal substitution.

Steps:

  1. Check the type of indeterminate form:

    • As x→0x \to 0, the numerator sin⁡3x→0\sin 3x \to 0
    • As x→0x \to 0, the denominator tan⁡2x→0\tan 2x \to 0
    • Hence it is a 00\frac{0}{0} indeterminate form

  2. Use equivalent infinitesimal substitution:

    • sin⁡3x∼3x\sin 3x \sim 3x
    • tan⁡2x∼2x\tan 2x \sim 2x

  3. Evaluate the limit after substitution: lim⁡x→0sin⁡3xtan⁡2x=lim⁡x→03x2x=lim⁡x→032=32\lim_{x \to 0} \frac{\sin 3x}{\tan 2x} = \lim_{x \to 0} \frac{3x}{2x} = \lim_{x \to 0} \frac{3}{2} = \frac{3}{2}

Answer: lim⁡x→0sin⁡3xtan⁡2x=32\lim_{x \to 0} \frac{\sin 3x}{\tan 2x} = \frac{3}{2}

Example 2

Find the limit lim⁡x→0ln⁡(1+x2)x2\lim_{x \to 0} \frac{\ln(1 + x^2)}{x^2}

equivalent infinitesimal

Approach: This is a 00\frac{0}{0} indeterminate form, so we can use equivalent infinitesimal substitution.

Steps:

  1. Check the type of indeterminate form:

    • As x→0x \to 0, the numerator ln⁡(1+x2)→0\ln(1 + x^2) \to 0
    • As x→0x \to 0, the denominator x2→0x^2 \to 0
    • Hence it is a 00\frac{0}{0} indeterminate form

  2. Use equivalent infinitesimal substitution:

    • ln⁡(1+x2)∼x2\ln(1 + x^2) \sim x^2

  3. Evaluate the limit after substitution: lim⁡x→0ln⁡(1+x2)x2=lim⁡x→0x2x2=lim⁡x→01=1\lim_{x \to 0} \frac{\ln(1 + x^2)}{x^2} = \lim_{x \to 0} \frac{x^2}{x^2} = \lim_{x \to 0} 1 = 1

Answer: lim⁡x→0ln⁡(1+x2)x2=1\lim_{x \to 0} \frac{\ln(1 + x^2)}{x^2} = 1

Practice Problems

Exercise 1

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

equivalent infinitesimal

Approach: This is a 00\frac{0}{0} indeterminate form, so we can use equivalent infinitesimal substitution.

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 sin⁡x→0\sin x \to 0
    • Hence it is a 00\frac{0}{0} indeterminate form

  2. Use equivalent infinitesimal substitution:

    • ex−1∼xe^x - 1 \sim x
    • sin⁡x∼x\sin x \sim x

  3. Evaluate the limit after substitution: lim⁡x→0ex−1sin⁡x=lim⁡x→0xx=lim⁡x→01=1\lim_{x \to 0} \frac{e^x - 1}{\sin x} = \lim_{x \to 0} \frac{x}{x} = \lim_{x \to 0} 1 = 1

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

Exercise 2

Find the limit lim⁡x→0tan⁡x−xx3\lim_{x \to 0} \frac{\tan x - x}{x^3}

equivalent infinitesimal

Approach: This is a 00\frac{0}{0} indeterminate form, so we need to use the higher-order equivalence relation.

Steps:

  1. Check the type of indeterminate form:

    • As x→0x \to 0, the numerator tan⁡x−x→0\tan x - x \to 0
    • As x→0x \to 0, the denominator x3→0x^3 \to 0
    • Hence it is a 00\frac{0}{0} indeterminate form

  2. Use the higher-order equivalence relation:

    • tan⁡x−x∼x33\tan x - x \sim \frac{x^3}{3}

  3. Evaluate the limit after substitution: lim⁡x→0tan⁡x−xx3=lim⁡x→0x33x3=lim⁡x→013=13\lim_{x \to 0} \frac{\tan x - x}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{3}}{x^3} = \lim_{x \to 0} \frac{1}{3} = \frac{1}{3}

Answer: lim⁡x→0tan⁡x−xx3=13\lim_{x \to 0} \frac{\tan x - x}{x^3} = \frac{1}{3}

Summary

Symbols Used in This Article

Symbol notation
SymbolTypePronunciation/ExplanationMeaning in This Article
lim⁡x→a\lim_{x \to a}Mathematical symbollimit as x approaches aThe limit as x→ax \to a
∼\simMathematical symboltilde (equivalence sign)Two infinitesimals are equivalent
α(x),β(x)\alpha(x), \beta(x)Greek letterAlpha/BetaInfinitesimals
f(x)f(x)Mathematical symbolf of xThe function compared in the limit
sin⁡x,tan⁡x\sin x, \tan xMathematical symbolsine/tangent of xTrigonometric functions
arcsin⁡x,arctan⁡x\arcsin x, \arctan xMathematical symbolarc sine/arc tangent of xInverse trigonometric functions
ln⁡(1+x)\ln(1 + x)Mathematical symbolnatural log of 1 plus xThe natural logarithm function
exe^xMathematical symbole to the xThe exponential function
(1+x)a(1 + x)^aMathematical symbol1 plus x to the aA power function
00\frac{0}{0}Mathematical symbolzero over zeroIndeterminate form
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