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 limf(x)=0\lim f(x) = 0 and limg(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 limf(x)=0\lim f(x) = 0 and limg(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 limf(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

  • limx0(x+x2)=0\lim_{x \to 0} (x + x^2) = 0 (the sum of two infinitesimals)
  • limx0xx2=0\lim_{x \to 0} x \cdot x^2 = 0 (the product of two infinitesimals)
  • limx0xsin1x=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 limf(x)=\lim f(x) = \infty and limg(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 limf(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 limf(x)=\lim f(x) = \infty and c0c \neq 0, then lim[cf(x)]=\lim [c \cdot f(x)] = \infty

Examples

  • limxxx2=\lim_{x \to \infty} x \cdot x^2 = \infty (the product of two infinities)
  • limx(x+sinx)=\lim_{x \to \infty} (x + \sin x) = \infty (the sum of an infinity and a bounded function)
  • limx3x=\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. 00 \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. 11^\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: limx2x24x2=limx2(x2)(x+2)x2=limx2(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: limxx2+3xx2+1=limx1+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 00 \cdot \infty Type

Methods:

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

Example: limx0+xlnx=limx0+lnx1x\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: limx(x2+xx)=limxxx2+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, 11^\infty Types

Methods:

  • Take logarithms to transform into the 00 \cdot \infty type

Example: limx0+xx=limx0+exlnx=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 limx0x2sin1x\lim_{x \to 0} x^2 \sin \frac{1}{x} and compute it.

Reference Answer(1 个标签)
limit operation rules

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

Detailed steps:

  1. As x0x \to 0, x20x^2 \to 0 (an infinitesimal)

  2. sin1x\sin \frac{1}{x} is bounded (1sin1x1-1 \leq \sin \frac{1}{x} \leq 1)

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

  4. Therefore limx0x2sin1x=0\lim_{x \to 0} x^2 \sin \frac{1}{x} = 0

Answer: The limit is 0.

Exercise 2

Compute the limit limx2x24x2\lim_{x \to 2} \frac{x^2 - 4}{x - 2} (of the 00\frac{0}{0} type).

Reference Answer(1 个标签)
limit operation rules

Idea: Factor and cancel the zero factor.

Detailed steps:

  1. x24x2=(x2)(x+2)x2=x+2\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x + 2 (when x2x \neq 2)

  2. limx2x24x2=limx2(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 limxx2+3x+1x2+2x\lim_{x \to \infty} \frac{x^2 + 3x + 1}{x^2 + 2x} (of the \frac{\infty}{\infty} type).

Reference Answer(1 个标签)
limit operation rules

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

Detailed 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. limxx2+3x+1x2+2x=limx1+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

符号类型读音/说明在本文中的含义
lim\lim数学符号limitThe limit of a function or sequence
\to数学符号tends toA variable tending to some value
\infty数学符号infinityInfinity
00数学符号zeroZero or an infinitesimal

中英对照

中文术语英文术语音标说明
不定式indeterminate form/ˌɪndɪˈtɜːmɪnət fɔːm/A limit form requiring special handling
无穷小infinitesimal/ˌɪnfɪnɪˈtesɪməl/A function or sequence whose limit is 0
无穷大infinity/ɪnˈfɪnɪti/A function or sequence whose limit is infinity
有界函数bounded function/ˈbaʊndɪd ˈfʌŋkʃən/A function whose values stay within some range
因式分解factorization/ˌfæktəraɪˈzeɪʃən/Decomposing a polynomial into a product of factors
有理化rationalization/ˌræʃənəlaɪˈzeɪʃən/Eliminating radicals from the numerator or denominator