Kuzee
The World of Limits in Advanced Mathematics

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)=xsin⁡1xg(x) = x \sin \frac{1}{x} is squeezed between the red dashed line f(x)=−∣x∣f(x) = -|x| and the green dashed line h(x)=∣x∣h(x) = |x|. As x→0x \to 0, both boundary curves tend to 00, so the squeezed function g(x)g(x) must also tend to 00.

Basic Principle

The squeeze theorem

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

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

and:

lim⁡x→af(x)=lim⁡x→ah(x)=L\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L

then:

lim⁡x→ag(x)=L\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\sin x, cos⁡x\cos x
  • Functions involving xsin⁡1xx \sin \frac{1}{x}
  • Functions involving xcos⁡1xx \cos \frac{1}{x}

Construction Techniques

Common inequalities for the squeeze theorem

1. Use the boundedness of trigonometric functions

−1≤sin⁡x≤1-1 \leq \sin x \leq 1 −1≤cos⁡x≤1-1 \leq \cos x \leq 1

2. Use absolute values

∣x∣≥0\vert x \vert \geq 0

3. Use basic inequalities

x2≥0x^2 \geq 0 ex>0e^x > 0

Worked Examples

Example 1

Find the limit lim⁡x→0xsin⁡1x\lim_{x \to 0} x \sin \frac{1}{x}

squeeze theorem

Approach: This limit is hard to compute directly, so we can use the squeeze theorem.

Steps:

  1. Analyze the features of the function:

    • As x→0x \to 0, we have x→0x \to 0
    • sin⁡1x\sin \frac{1}{x} oscillates between [−1,1][-1, 1]
    • It cannot be computed directly

  2. Construct the inequality:

    • Since −1≤sin⁡1x≤1-1 \leq \sin \frac{1}{x} \leq 1
    • We have −∣x∣≤xsin⁡1x≤∣x∣-|x| \leq x \sin \frac{1}{x} \leq |x|

  3. 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 \to 0} |x| = 0

  4. Apply the squeeze theorem:

    • Since −∣x∣≤xsin⁡1x≤∣x∣-|x| \leq x \sin \frac{1}{x} \leq |x|
    • And lim⁡x→0(−∣x∣)=lim⁡x→0∣x∣=0\lim_{x \to 0} (-|x|) = \lim_{x \to 0} |x| = 0
    • Therefore lim⁡x→0xsin⁡1x=0\lim_{x \to 0} x \sin \frac{1}{x} = 0

Answer: lim⁡x→0xsin⁡1x=0\lim_{x \to 0} x \sin \frac{1}{x} = 0

Example 2

Find the limit lim⁡x→∞sin⁡xx\lim_{x \to \infty} \frac{\sin x}{x}

squeeze theorem

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

Steps:

  1. Analyze the features of the function:

    • As x→∞x \to \infty, the denominator x→∞x \to \infty
    • The numerator sin⁡x\sin x oscillates between [−1,1][-1, 1]

  2. Construct the inequality:

    • Since −1≤sin⁡x≤1-1 \leq \sin x \leq 1
    • We have −1x≤sin⁡xx≤1x-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x}

  3. Find the limits of the boundary functions:

    • lim⁡x→∞(−1x)=0\lim_{x \to \infty} (-\frac{1}{x}) = 0
    • lim⁡x→∞1x=0\lim_{x \to \infty} \frac{1}{x} = 0

  4. Apply the squeeze theorem:

    • Since −1x≤sin⁡xx≤1x-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x}
    • And lim⁡x→∞(−1x)=lim⁡x→∞1x=0\lim_{x \to \infty} (-\frac{1}{x}) = \lim_{x \to \infty} \frac{1}{x} = 0
    • Therefore lim⁡x→∞sin⁡xx=0\lim_{x \to \infty} \frac{\sin x}{x} = 0

Answer: lim⁡x→∞sin⁡xx=0\lim_{x \to \infty} \frac{\sin x}{x} = 0

Practice Problems

Exercise 1

Find the limit lim⁡x→0x2cos⁡1x\lim_{x \to 0} x^2 \cos \frac{1}{x}

squeeze theorem

Approach: This limit is hard to compute directly, so we can use the squeeze theorem.

Steps:

  1. Analyze the features of the function:

    • As x→0x \to 0, we have x2→0x^2 \to 0
    • cos⁡1x\cos \frac{1}{x} oscillates between [−1,1][-1, 1]

  2. Construct the inequality:

    • Since −1≤cos⁡1x≤1-1 \leq \cos \frac{1}{x} \leq 1
    • We have −x2≤x2cos⁡1x≤x2-x^2 \leq x^2 \cos \frac{1}{x} \leq x^2

  3. Find the limits of the boundary functions:

    • lim⁡x→0(−x2)=0\lim_{x \to 0} (-x^2) = 0
    • lim⁡x→0x2=0\lim_{x \to 0} x^2 = 0

  4. Apply the squeeze theorem:

    • Since −x2≤x2cos⁡1x≤x2-x^2 \leq x^2 \cos \frac{1}{x} \leq x^2
    • And lim⁡x→0(−x2)=lim⁡x→0x2=0\lim_{x \to 0} (-x^2) = \lim_{x \to 0} x^2 = 0
    • Therefore lim⁡x→0x2cos⁡1x=0\lim_{x \to 0} x^2 \cos \frac{1}{x} = 0

Answer: lim⁡x→0x2cos⁡1x=0\lim_{x \to 0} x^2 \cos \frac{1}{x} = 0

Exercise 2

Find the limit lim⁡x→0x2sin⁡1xx\lim_{x \to 0} \frac{x^2 \sin \frac{1}{x}}{x}

squeeze theorem

Approach: This is a 00\frac{0}{0} indeterminate form, so we can use the squeeze theorem.

Steps:

  1. Simplify the expression: x2sin⁡1xx=xsin⁡1x\frac{x^2 \sin \frac{1}{x}}{x} = x \sin \frac{1}{x}

  2. Analyze the features of the function:

    • As x→0x \to 0, we have x→0x \to 0
    • sin⁡1x\sin \frac{1}{x} oscillates between [−1,1][-1, 1]

  3. Construct the inequality:

    • Since −1≤sin⁡1x≤1-1 \leq \sin \frac{1}{x} \leq 1
    • We have −∣x∣≤xsin⁡1x≤∣x∣-|x| \leq x \sin \frac{1}{x} \leq |x|

  4. 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 \to 0} |x| = 0

  5. Apply the squeeze theorem:

    • Since −∣x∣≤xsin⁡1x≤∣x∣-|x| \leq x \sin \frac{1}{x} \leq |x|
    • And lim⁡x→0(−∣x∣)=lim⁡x→0∣x∣=0\lim_{x \to 0} (-|x|) = \lim_{x \to 0} |x| = 0
    • Therefore lim⁡x→0xsin⁡1x=0\lim_{x \to 0} x \sin \frac{1}{x} = 0

Answer: lim⁡x→0x2sin⁡1xx=0\lim_{x \to 0} \frac{x^2 \sin \frac{1}{x}}{x} = 0

Summary

Symbols Used in This Article

Symbol notation
SymbolTypePronunciation/ExplanationMeaning in This Article
lim⁡x→a\lim_{x \to a}Mathematical symbollimit as x approaches aThe limit as x→ax \to a
f(x),g(x),h(x)f(x), g(x), h(x)Mathematical symbolf/g/h of xLower bound, target, and upper bound functions
LLMathematical symbolLThe common limit of the upper and lower bounds
∣x∣\lvert x\rvertMathematical symbolabsolute value of xThe absolute value of xx
sin⁡x,cos⁡x\sin x, \cos xMathematical symbolsine/cosine of xTrigonometric functions
1x\frac{1}{x}Mathematical symbolone over xThe reciprocal function
∞\inftyMathematical symbolinfinityInfinity
≤\leqMathematical symbolless than or equal toLess than or equal to
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