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The World of Limits in Advanced Mathematics

Special Cases

In limit operations, there are some special cases that need particular attention, including operations involving infinitesimals, infinities, and indeterminate forms.

Operations with Infinitesimals

Operation rules for infinitesimals
  1. The sum of finitely many infinitesimals is still an infinitesimal

If lim⁡f(x)=0\lim f(x) = 0 and lim⁡g(x)=0\lim g(x) = 0, then lim⁡[f(x)+g(x)]=0\lim [f(x) + g(x)] = 0

  1. The product of finitely many infinitesimals is still an infinitesimal

    If lim⁡f(x)=0\lim f(x) = 0 and lim⁡g(x)=0\lim g(x) = 0, then lim⁡[f(x)⋅g(x)]=0\lim [f(x) \cdot g(x)] = 0

  2. The product of a bounded function and an infinitesimal is still an infinitesimal

    If lim⁡f(x)=0\lim f(x) = 0 and g(x)g(x) is bounded, then lim⁡[f(x)⋅g(x)]=0\lim [f(x) \cdot g(x)] = 0

Examples

  • lim⁡x→0(x+x2)=0\lim_{x \to 0} (x + x^2) = 0 (the sum of two infinitesimals)
  • lim⁡x→0x⋅x2=0\lim_{x \to 0} x \cdot x^2 = 0 (the product of two infinitesimals)
  • lim⁡x→0xsin⁡1x=0\lim_{x \to 0} x \sin \frac{1}{x} = 0 (the product of a bounded function and an infinitesimal)

Operations with Infinities

Operation rules for infinities
  1. The product of finitely many infinities is still infinity

If lim⁡f(x)=∞\lim f(x) = \infty and lim⁡g(x)=∞\lim g(x) = \infty, then lim⁡[f(x)⋅g(x)]=∞\lim [f(x) \cdot g(x)] = \infty

  1. The sum of an infinity and a bounded function is still infinity

    If lim⁡f(x)=∞\lim f(x) = \infty and g(x)g(x) is bounded, then lim⁡[f(x)+g(x)]=∞\lim [f(x) + g(x)] = \infty

  2. The product of an infinity and a nonzero constant is still infinity

    If lim⁡f(x)=∞\lim f(x) = \infty and c≠0c \neq 0, then lim⁡[c⋅f(x)]=∞\lim [c \cdot f(x)] = \infty

Examples

  • lim⁡x→∞x⋅x2=∞\lim_{x \to \infty} x \cdot x^2 = \infty (the product of two infinities)
  • lim⁡x→∞(x+sin⁡x)=∞\lim_{x \to \infty} (x + \sin x) = \infty (the sum of an infinity and a bounded function)
  • lim⁡x→∞3x=∞\lim_{x \to \infty} 3x = \infty (the product of an infinity and a nonzero constant)

Indeterminate Forms

In limit computation, the following cases are called indeterminate forms and require special handling:

The seven indeterminate forms

  1. 00\frac{0}{0} type: both the numerator and the denominator tend to 0

  2. ∞∞\frac{\infty}{\infty} type: both the numerator and the denominator tend to infinity

  3. 0⋅∞0 \cdot \infty type: one tends to 0, the other tends to infinity

  4. ∞−∞\infty - \infty type: both tend to infinity

  5. 000^0 type: the base tends to 0, the exponent tends to 0

  6. ∞0\infty^0 type: the base tends to infinity, the exponent tends to 0

  7. 1∞1^\infty type: the base tends to 1, the exponent tends to infinity

How to Handle Indeterminate Forms

1. The 00\frac{0}{0} Type

Methods:

  • Factorization
  • Rationalization
  • Equivalent infinitesimal substitution
  • L’Hôpital’s rule

Example: lim⁡x→2x2−4x−2=lim⁡x→2(x−2)(x+2)x−2=lim⁡x→2(x+2)=4\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4

2. The ∞∞\frac{\infty}{\infty} Type

Methods:

  • Divide both the numerator and the denominator by the highest-degree term
  • L’Hôpital’s rule

Example: lim⁡x→∞x2+3xx2+1=lim⁡x→∞1+3x1+1x2=1\lim_{x \to \infty} \frac{x^2 + 3x}{x^2 + 1} = \lim_{x \to \infty} \frac{1 + \frac{3}{x}}{1 + \frac{1}{x^2}} = 1

3. The 0⋅∞0 \cdot \infty Type

Methods:

  • Transform into the 00\frac{0}{0} or ∞∞\frac{\infty}{\infty} type

Example: lim⁡x→0+xln⁡x=lim⁡x→0+ln⁡x1x\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x}} (transformed into the ∞∞\frac{\infty}{\infty} type)

4. The ∞−∞\infty - \infty Type

Methods:

  • Combine fractions
  • Rationalize
  • Factor out a common factor

Example: lim⁡x→∞(x2+x−x)=lim⁡x→∞xx2+x+x=12\lim_{x \to \infty} (\sqrt{x^2 + x} - x) = \lim_{x \to \infty} \frac{x}{\sqrt{x^2 + x} + x} = \frac{1}{2}

5. The 000^0, ∞0\infty^0, 1∞1^\infty Types

Methods:

  • Take logarithms to transform into the 0⋅∞0 \cdot \infty type

Example: lim⁡x→0+xx=lim⁡x→0+exln⁡x=e0=1\lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{x \ln x} = e^0 = 1

Practice Problems

Exercise 1

Identify the type of lim⁡x→0x2sin⁡1x\lim_{x \to 0} x^2 \sin \frac{1}{x} and compute it.

limit operation rules

Approach: This is the product of a bounded function and an infinitesimal.

Steps:

  1. As x→0x \to 0, x2→0x^2 \to 0 (an infinitesimal)

  2. sin⁡1x\sin \frac{1}{x} is bounded (−1≤sin⁡1x≤1-1 \leq \sin \frac{1}{x} \leq 1)

  3. The product of a bounded function and an infinitesimal is still an infinitesimal

  4. Therefore lim⁡x→0x2sin⁡1x=0\lim_{x \to 0} x^2 \sin \frac{1}{x} = 0

Answer: The limit is 0.

Exercise 2

Compute the limit lim⁡x→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2} (of the 00\frac{0}{0} type).

limit operation rules

Approach: Factor and cancel the zero factor.

Steps:

  1. x2−4x−2=(x−2)(x+2)x−2=x+2\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x + 2 (when x≠2x \neq 2)

  2. lim⁡x→2x2−4x−2=lim⁡x→2(x+2)=4\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} (x + 2) = 4

Answer: The limit is 4.

Exercise 3

Compute the limit lim⁡x→∞x2+3x+1x2+2x\lim_{x \to \infty} \frac{x^2 + 3x + 1}{x^2 + 2x} (of the ∞∞\frac{\infty}{\infty} type).

limit operation rules

Approach: Divide both the numerator and the denominator by the highest-degree term.

Steps:

  1. x2+3x+1x2+2x=1+3x+1x21+2x\frac{x^2 + 3x + 1}{x^2 + 2x} = \frac{1 + \frac{3}{x} + \frac{1}{x^2}}{1 + \frac{2}{x}}

  2. lim⁡x→∞x2+3x+1x2+2x=lim⁡x→∞1+3x+1x21+2x\lim_{x \to \infty} \frac{x^2 + 3x + 1}{x^2 + 2x} = \lim_{x \to \infty} \frac{1 + \frac{3}{x} + \frac{1}{x^2}}{1 + \frac{2}{x}}

  3. =1+0+01+0=1= \frac{1 + 0 + 0}{1 + 0} = 1

Answer: The limit is 1.

Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
lim⁡\limMathematical symbollimitThe limit of a function or sequence
→\toMathematical symboltends toA variable tending to some value
∞\inftyMathematical symbolinfinityInfinity
00Mathematical symbolzeroZero or an infinitesimal
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Limits of Composite Functions

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