This is a beta course, so its structure, chapters, and examples may continue to change.
The Squeeze Theorem
The squeeze theorem is an important method for evaluating limits, especially for limits that are hard to compute directly. By constructing two functions with known limits to “squeeze” the target function, we can determine its limit.
Geometric Interpretation As shown in the figure, the blue curve g ( x ) = x sin 1 x g(x) = x \sin \frac{1}{x} g ( x ) = x sin x 1 f ( x ) = − ∣ x ∣ f(x) = -|x| f ( x ) = − ∣ x ∣ h ( x ) = ∣ x ∣ h(x) = |x| h ( x ) = ∣ x ∣ x → 0 x \to 0 x → 0 0 0 0 g ( x ) g(x) g ( x ) 0 0 0
Basic Principle 定义
定义是对概念、术语或对象含义的精确描述。理解定义是学习任何知识领域的基础，每个概念都有其明确的定义。
If for all x x x
f ( x ) ≤ g ( x ) ≤ h ( x ) f(x) \leq g(x) \leq h(x) f ( x ) ≤ g ( x ) ≤ h ( x )
and:
lim x → a f ( x ) = lim x → a h ( x ) = L \lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L lim x → a f ( x ) = lim x → a h ( x ) = L
then:
lim x → a g ( x ) = L \lim_{x \to a} g(x) = L lim x → a g ( x ) = L
Applicable Conditions 1. Direct Computation Is Difficult
The function has a complicated form
No standard method applies
It involves trigonometric, exponential, or other functions
2. Inequalities Can Be Constructed
We can find upper and lower bounds
The limits of the upper and lower bounds are equal
3. Common Cases
Functions involving sin x \sin x sin x cos x \cos x cos x
Functions involving x sin 1 x x \sin \frac{1}{x} x sin x 1
Functions involving x cos 1 x x \cos \frac{1}{x} x cos x 1
Construction Techniques Worked Examples Example 1
Find the limit lim x → 0 x sin 1 x \lim_{x \to 0} x \sin \frac{1}{x} lim x → 0 x sin x 1
Reference Answer
squeeze theorem
Idea : This limit is hard to compute directly, so we can use the squeeze theorem.
Detailed steps :
Analyze the features of the function:
As x → 0 x \to 0 x → 0 x → 0 x \to 0 x → 0
sin 1 x \sin \frac{1}{x} sin x 1 [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] It cannot be computed directly
Construct the inequality:
Since − 1 ≤ sin 1 x ≤ 1 -1 \leq \sin \frac{1}{x} \leq 1 − 1 ≤ sin x 1 ≤ 1
We have − ∣ x ∣ ≤ x sin 1 x ≤ ∣ x ∣ -|x| \leq x \sin \frac{1}{x} \leq |x| − ∣ x ∣ ≤ x sin x 1 ≤ ∣ x ∣
Find the limits of the boundary functions:
lim x → 0 ( − ∣ x ∣ ) = 0 \lim_{x \to 0} (-|x|) = 0 lim x → 0 ( − ∣ x ∣ ) = 0 lim x → 0 ∣ x ∣ = 0 \lim_{x \to 0} |x| = 0 lim x → 0 ∣ x ∣ = 0
Apply the squeeze theorem:
Since − ∣ x ∣ ≤ x sin 1 x ≤ ∣ x ∣ -|x| \leq x \sin \frac{1}{x} \leq |x| − ∣ x ∣ ≤ x sin x 1 ≤ ∣ x ∣
And lim x → 0 ( − ∣ x ∣ ) = lim x → 0 ∣ x ∣ = 0 \lim_{x \to 0} (-|x|) = \lim_{x \to 0} |x| = 0 lim x → 0 ( − ∣ x ∣ ) = lim x → 0 ∣ x ∣ = 0
Therefore lim x → 0 x sin 1 x = 0 \lim_{x \to 0} x \sin \frac{1}{x} = 0 lim x → 0 x sin x 1 = 0
Answer : lim x → 0 x sin 1 x = 0 \lim_{x \to 0} x \sin \frac{1}{x} = 0 lim x → 0 x sin x 1 = 0
Example 2
Find the limit lim x → ∞ sin x x \lim_{x \to \infty} \frac{\sin x}{x} lim x → ∞ x s i n x
Reference Answer
squeeze theorem
Idea : This is a 0 ∞ \frac{0}{\infty} ∞ 0
Detailed steps :
Analyze the features of the function:
As x → ∞ x \to \infty x → ∞ x → ∞ x \to \infty x → ∞
The numerator sin x \sin x sin x [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ]
Construct the inequality:
Since − 1 ≤ sin x ≤ 1 -1 \leq \sin x \leq 1 − 1 ≤ sin x ≤ 1
We have − 1 x ≤ sin x x ≤ 1 x -\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} − x 1 ≤ x s i n x ≤ x 1
Find the limits of the boundary functions:
lim x → ∞ ( − 1 x ) = 0 \lim_{x \to \infty} (-\frac{1}{x}) = 0 lim x → ∞ ( − x 1 ) = 0 lim x → ∞ 1 x = 0 \lim_{x \to \infty} \frac{1}{x} = 0 lim x → ∞ x 1 = 0
Apply the squeeze theorem:
Since − 1 x ≤ sin x x ≤ 1 x -\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} − x 1 ≤ x s i n x ≤ x 1
And lim x → ∞ ( − 1 x ) = lim x → ∞ 1 x = 0 \lim_{x \to \infty} (-\frac{1}{x}) = \lim_{x \to \infty} \frac{1}{x} = 0 lim x → ∞ ( − x 1 ) = lim x → ∞ x 1 = 0
Therefore lim x → ∞ sin x x = 0 \lim_{x \to \infty} \frac{\sin x}{x} = 0 lim x → ∞ x s i n x = 0
Answer : lim x → ∞ sin x x = 0 \lim_{x \to \infty} \frac{\sin x}{x} = 0 lim x → ∞ x s i n x = 0
Practice Problems Exercise 1
Find the limit lim x → 0 x 2 cos 1 x \lim_{x \to 0} x^2 \cos \frac{1}{x} lim x → 0 x 2 cos x 1
Reference Answer
squeeze theorem
Idea : This limit is hard to compute directly, so we can use the squeeze theorem.
Detailed steps :
Analyze the features of the function:
As x → 0 x \to 0 x → 0 x 2 → 0 x^2 \to 0 x 2 → 0
cos 1 x \cos \frac{1}{x} cos x 1 [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ]
Construct the inequality:
Since − 1 ≤ cos 1 x ≤ 1 -1 \leq \cos \frac{1}{x} \leq 1 − 1 ≤ cos x 1 ≤ 1
We have − x 2 ≤ x 2 cos 1 x ≤ x 2 -x^2 \leq x^2 \cos \frac{1}{x} \leq x^2 − x 2 ≤ x 2 cos x 1 ≤ x 2
Find the limits of the boundary functions:
lim x → 0 ( − x 2 ) = 0 \lim_{x \to 0} (-x^2) = 0 lim x → 0 ( − x 2 ) = 0 lim x → 0 x 2 = 0 \lim_{x \to 0} x^2 = 0 lim x → 0 x 2 = 0
Apply the squeeze theorem:
Since − x 2 ≤ x 2 cos 1 x ≤ x 2 -x^2 \leq x^2 \cos \frac{1}{x} \leq x^2 − x 2 ≤ x 2 cos x 1 ≤ x 2
And lim x → 0 ( − x 2 ) = lim x → 0 x 2 = 0 \lim_{x \to 0} (-x^2) = \lim_{x \to 0} x^2 = 0 lim x → 0 ( − x 2 ) = lim x → 0 x 2 = 0
Therefore lim x → 0 x 2 cos 1 x = 0 \lim_{x \to 0} x^2 \cos \frac{1}{x} = 0 lim x → 0 x 2 cos x 1 = 0
Answer : lim x → 0 x 2 cos 1 x = 0 \lim_{x \to 0} x^2 \cos \frac{1}{x} = 0 lim x → 0 x 2 cos x 1 = 0
Exercise 2
Find the limit lim x → 0 x 2 sin 1 x x \lim_{x \to 0} \frac{x^2 \sin \frac{1}{x}}{x} lim x → 0 x x 2 s i n x 1
Reference Answer
squeeze theorem
Idea : This is a 0 0 \frac{0}{0} 0 0
Detailed steps :
Simplify the expression: x 2 sin 1 x x = x sin 1 x \frac{x^2 \sin \frac{1}{x}}{x} = x \sin \frac{1}{x} x x 2 s i n x 1 = x sin x 1
Analyze the features of the function:
As x → 0 x \to 0 x → 0 x → 0 x \to 0 x → 0
sin 1 x \sin \frac{1}{x} sin x 1 [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ]
Construct the inequality:
Since − 1 ≤ sin 1 x ≤ 1 -1 \leq \sin \frac{1}{x} \leq 1 − 1 ≤ sin x 1 ≤ 1
We have − ∣ x ∣ ≤ x sin 1 x ≤ ∣ x ∣ -|x| \leq x \sin \frac{1}{x} \leq |x| − ∣ x ∣ ≤ x sin x 1 ≤ ∣ x ∣
Find the limits of the boundary functions:
lim x → 0 ( − ∣ x ∣ ) = 0 \lim_{x \to 0} (-|x|) = 0 lim x → 0 ( − ∣ x ∣ ) = 0 lim x → 0 ∣ x ∣ = 0 \lim_{x \to 0} |x| = 0 lim x → 0 ∣ x ∣ = 0
Apply the squeeze theorem:
Since − ∣ x ∣ ≤ x sin 1 x ≤ ∣ x ∣ -|x| \leq x \sin \frac{1}{x} \leq |x| − ∣ x ∣ ≤ x sin x 1 ≤ ∣ x ∣
And lim x → 0 ( − ∣ x ∣ ) = lim x → 0 ∣ x ∣ = 0 \lim_{x \to 0} (-|x|) = \lim_{x \to 0} |x| = 0 lim x → 0 ( − ∣ x ∣ ) = lim x → 0 ∣ x ∣ = 0
Therefore lim x → 0 x sin 1 x = 0 \lim_{x \to 0} x \sin \frac{1}{x} = 0 lim x → 0 x sin x 1 = 0
Answer : lim x → 0 x 2 sin 1 x x = 0 \lim_{x \to 0} \frac{x^2 \sin \frac{1}{x}}{x} = 0 lim x → 0 x x 2 s i n x 1 = 0
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} lim x → a 数学符号 limit as x approaches a The limit as x → a x \to a x → a f ( x ) , g ( x ) , h ( x ) f(x), g(x), h(x) f ( x ) , g ( x ) , h ( x ) 数学符号 f/g/h of x Lower bound, target, and upper bound functions L L L 数学符号 L The common limit of the upper and lower bounds $ x $ 数学符号 sin x , cos x \sin x, \cos x sin x , cos x 数学符号 sine/cosine of x Trigonometric functions 1 x \frac{1}{x} x 1 数学符号 one over x The reciprocal function ∞ \infty ∞ 数学符号 infinity Infinity ≤ \leq ≤ 数学符号 less than or equal to Less than or equal to
中英对照
中文术语 英文术语 音标 说明 夹逼准则 squeeze theorem /skwiːz ˈθɪərəm/ Finding a limit by squeezing between upper and lower bounds 去心邻域 punctured neighborhood /ˈpʌŋktʃəd ˈneɪbəhʊd/ A neighborhood without its center point 上界 upper bound /ˈʌpə baʊnd/ A number no smaller than all function values 下界 lower bound /ˈləʊə baʊnd/ A number no larger than all function values 有界性 boundedness /ˈbaʊndɪdnəs/ The property that function values stay bounded 振荡 oscillate /ˈɒsɪleɪt/ To swing back and forth within an interval 绝对值 absolute value /ˈæbsəluːt ˈvæljuː/ The nonnegative value of a number
