Limits of Trigonometric Functions

Limits of trigonometric functions are important formulas in limit theory and are frequently used in limit computation.

Limit of the Cosine Function

Limit of the cosine function

lim⁡x→01−cos⁡xx2=12\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}

Equivalent infinitesimal: as x→0x \to 0, 1−cos⁡x∼x221 - \cos x \sim \frac{x^2}{2}

Limit of the Tangent Function

Limit of the tangent function

lim⁡x→0tan⁡xx=1\lim_{x \to 0} \frac{\tan x}{x} = 1

Equivalent infinitesimal: as x→0x \to 0, tan⁡x∼x\tan x \sim x

Idea of the proof: use the first important limit: lim⁡x→0tan⁡xx=lim⁡x→0sin⁡xxcos⁡x=11=1\lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x \cos x} = \frac{1}{1} = 1

Basic Form

lim⁡x→0tan⁡xx=1\lim_{x \to 0} \frac{\tan x}{x} = 1

Idea of the Proof

Use the first important limit: lim⁡x→0tan⁡xx=lim⁡x→0sin⁡xxcos⁡x=11=1\lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x \cos x} = \frac{1}{1} = 1

Equivalent Infinitesimal

Using this important limit, we obtain an important equivalent infinitesimal:

  • As x→0x \to 0, tan⁡x∼x\tan x \sim x

Worked Example

Find lim⁡x→0tan⁡2xx\lim_{x \to 0} \frac{\tan 2x}{x}

Solution: lim⁡x→0tan⁡2xx=lim⁡x→02⋅tan⁡2x2x=2⋅1=2\lim_{x \to 0} \frac{\tan 2x}{x} = \lim_{x \to 0} 2 \cdot \frac{\tan 2x}{2x} = 2 \cdot 1 = 2


Practice Problems

Exercise 1

Find the limit lim⁡x→0tan⁡2xx\lim_{x \to 0} \frac{\tan 2x}{x}.

Reference Answer(1 个标签)
trigonometric limit

Idea: Use the tangent limit and a change of variable.

Detailed steps:

  1. lim⁡x→0tan⁡2xx=lim⁡x→02⋅tan⁡2x2x=2⋅1=2\lim_{x \to 0} \frac{\tan 2x}{x} = \lim_{x \to 0} 2 \cdot \frac{\tan 2x}{2x} = 2 \cdot 1 = 2

Answer: The limit is 2.

Exercise 2

Find the limit lim⁡x→01−cos⁡2xx2\lim_{x \to 0} \frac{1 - \cos 2x}{x^2}.

Reference Answer(1 个标签)
trigonometric limit

Idea: Use the cosine limit and a change of variable.

Detailed steps:

  1. lim⁡x→01−cos⁡2xx2=lim⁡x→04⋅1−cos⁡2x(2x)2=4⋅12=2\lim_{x \to 0} \frac{1 - \cos 2x}{x^2} = \lim_{x \to 0} 4 \cdot \frac{1 - \cos 2x}{(2x)^2} = 4 \cdot \frac{1}{2} = 2

Answer: The limit is 2.

Exercise 3

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

Reference Answer(1 个标签)
trigonometric limit

Idea: Use equivalent infinitesimal substitution.

Detailed steps:

  1. As x→0x \to 0, tan⁡3x∼3x\tan 3x \sim 3x and sin⁡2x∼2x\sin 2x \sim 2x

  2. lim⁡x→0tan⁡3xsin⁡2x=lim⁡x→03x2x=32\lim_{x \to 0} \frac{\tan 3x}{\sin 2x} = \lim_{x \to 0} \frac{3x}{2x} = \frac{3}{2}

Answer: The limit is 32\frac{3}{2}.


Summary

Symbols Used in This Article

符号类型读音/说明在本文中的含义
sin⁡\sin数学符号sine functionA trigonometric function
cos⁡\cos数学符号cosine functionA trigonometric function
tan⁡\tan数学符号tangent functionA trigonometric function
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
∼\sim数学符号equivalence signDenotes equivalent infinitesimals

中英对照

中文术语英文术语音标说明
三角函数trigonometric function/ˌtrɪɡənəˈmetrɪk ˈfʌŋkʃən/Functions related to angles
正弦函数sine function/saɪn ˈfʌŋkʃən/A trigonometric function
余弦函数cosine function/ˈkəʊsaɪn ˈfʌŋkʃən/A trigonometric function
正切函数tangent function/ˈtændʒənt ˈfʌŋkʃən/A trigonometric function
等价无穷小equivalent infinitesimal/ɪˈkwɪvələnt ˌɪnfɪnɪˈtesɪməl/Two infinitesimals whose ratio tends to 1
变量代换variable substitution/ˈveəriəbəl ˌsʌbstɪˈtjuːʃən/Replacing the original variable with a new one