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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 $g(x) = x \sin \frac{1}{x}$ is squeezed between the red dashed line $f(x) = -|x|$ and the green dashed line $h(x) = |x|$ . As $x \to 0$ , both boundary curves tend to $0$ , so the squeezed function $g(x)$ must also tend to $0$ .

Basic Principle

The squeeze theorem

If for all $x$ (in some punctured neighborhood), we have:

$f(x) \leq g(x) \leq h(x)$

and:

$\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$

then:

$\lim_{x \to a} g(x) = L$

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 $\sin x$ , $\cos x$
Functions involving $x \sin \frac{1}{x}$
Functions involving $x \cos \frac{1}{x}$

Construction Techniques

Common inequalities for the squeeze theorem

1. Use the boundedness of trigonometric functions

$-1 \leq \sin x \leq 1$ $-1 \leq \cos x \leq 1$

2. Use absolute values

$\vert x \vert \geq 0$

3. Use basic inequalities

$x^2 \geq 0$ $e^x > 0$

Worked Examples

Example 1

Find the limit $\lim_{x \to 0} x \sin \frac{1}{x}$

Reference Answer(1 个标签)

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 $x \to 0$ , we have $x \to 0$
- $\sin \frac{1}{x}$ oscillates between $[-1, 1]$
- It cannot be computed directly
Construct the inequality:
- Since $-1 \leq \sin \frac{1}{x} \leq 1$
- We have $-|x| \leq x \sin \frac{1}{x} \leq |x|$
Find the limits of the boundary functions:
- $\lim_{x \to 0} (-|x|) = 0$
- $\lim_{x \to 0} |x| = 0$
Apply the squeeze theorem:
- Since $-|x| \leq x \sin \frac{1}{x} \leq |x|$
- And $\lim_{x \to 0} (-|x|) = \lim_{x \to 0} |x| = 0$
- Therefore $\lim_{x \to 0} x \sin \frac{1}{x} = 0$

Answer: $\lim_{x \to 0} x \sin \frac{1}{x} = 0$

Example 2

Find the limit $\lim_{x \to \infty} \frac{\sin x}{x}$

Reference Answer(1 个标签)

squeeze theorem

Idea: This is a $\frac{0}{\infty}$ type limit, so we can use the squeeze theorem.

Detailed steps:

Analyze the features of the function:
- As $x \to \infty$ , the denominator $x \to \infty$
- The numerator $\sin x$ oscillates between $[-1, 1]$
Construct the inequality:
- Since $-1 \leq \sin x \leq 1$
- We have $-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x}$
Find the limits of the boundary functions:
- $\lim_{x \to \infty} (-\frac{1}{x}) = 0$
- $\lim_{x \to \infty} \frac{1}{x} = 0$
Apply the squeeze theorem:
- Since $-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x}$
- And $\lim_{x \to \infty} (-\frac{1}{x}) = \lim_{x \to \infty} \frac{1}{x} = 0$
- Therefore $\lim_{x \to \infty} \frac{\sin x}{x} = 0$

Answer: $\lim_{x \to \infty} \frac{\sin x}{x} = 0$

Practice Problems

Exercise 1

Find the limit $\lim_{x \to 0} x^2 \cos \frac{1}{x}$

Reference Answer(1 个标签)

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 $x \to 0$ , we have $x^2 \to 0$
- $\cos \frac{1}{x}$ oscillates between $[-1, 1]$
Construct the inequality:
- Since $-1 \leq \cos \frac{1}{x} \leq 1$
- We have $-x^2 \leq x^2 \cos \frac{1}{x} \leq x^2$
Find the limits of the boundary functions:
- $\lim_{x \to 0} (-x^2) = 0$
- $\lim_{x \to 0} x^2 = 0$
Apply the squeeze theorem:
- Since $-x^2 \leq x^2 \cos \frac{1}{x} \leq x^2$
- And $\lim_{x \to 0} (-x^2) = \lim_{x \to 0} x^2 = 0$
- Therefore $\lim_{x \to 0} x^2 \cos \frac{1}{x} = 0$

Answer: $\lim_{x \to 0} x^2 \cos \frac{1}{x} = 0$

Exercise 2

Find the limit $\lim_{x \to 0} \frac{x^2 \sin \frac{1}{x}}{x}$

Reference Answer(1 个标签)

squeeze theorem

Idea: This is a $\frac{0}{0}$ indeterminate form, so we can use the squeeze theorem.

Detailed steps:

Simplify the expression: $\frac{x^2 \sin \frac{1}{x}}{x} = x \sin \frac{1}{x}$
Analyze the features of the function:
- As $x \to 0$ , we have $x \to 0$
- $\sin \frac{1}{x}$ oscillates between $[-1, 1]$
Construct the inequality:
- Since $-1 \leq \sin \frac{1}{x} \leq 1$
- We have $-|x| \leq x \sin \frac{1}{x} \leq |x|$
Find the limits of the boundary functions:
- $\lim_{x \to 0} (-|x|) = 0$
- $\lim_{x \to 0} |x| = 0$
Apply the squeeze theorem:
- Since $-|x| \leq x \sin \frac{1}{x} \leq |x|$
- And $\lim_{x \to 0} (-|x|) = \lim_{x \to 0} |x| = 0$
- Therefore $\lim_{x \to 0} x \sin \frac{1}{x} = 0$

Answer: $\lim_{x \to 0} \frac{x^2 \sin \frac{1}{x}}{x} = 0$

Summary

Symbols Used in This Article

符号	类型	读音/说明	在本文中的含义
$\lim_{x \to a}$	数学符号	limit as x approaches a	The limit as $x \to a$
$f(x), g(x), h(x)$	数学符号	f/g/h of x	Lower bound, target, and upper bound functions
$L$	数学符号	L	The common limit of the upper and lower bounds
$	x	$	数学符号
$\sin x, \cos x$	数学符号	sine/cosine of x	Trigonometric functions
$\frac{1}{x}$	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

Equivalent Infinitesimal Substitution

L'Hôpital's Rule