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The Squeeze
Theorem

Master the art of bounding a function between two simpler functions to determine limits that seem impossible to evaluate directly.

AP Calculus AB

What Is a Limit?

A limit describes the value a function \( f(x) \) approaches as the input \( x \) gets closer and closer to some number \( a \). We write this as:

Sometimes direct substitution or algebraic manipulation can find \( L \). But what about functions like \( x^2 \sin\!\left(\frac{1}{x}\right) \) near \( x = 0 \)? We need a different strategy.

The Intuition Behind Squeezing

Imagine you are walking between two friends down a narrowing hallway. As the walls close in, you are forced toward the same point they converge to. That is exactly how the Squeeze Theorem works.

Upper bound
A function \( h(x) \) that is always above (or equal to) the target function near \( a \).
Target
The function \( f(x) \) whose limit we want to find: it is trapped in between.
Lower bound
A function \( g(x) \) that is always below (or equal to) the target function near \( a \).

Visualizing the Squeeze

The Squeeze Theorem: Formal Statement

The theorem works because the bounding functions leave \( f(x) \) no room to go anywhere else. Both bounds converge to the same value, forcing \( f(x) \) to follow.

The Theorem at a Glance

Squeeze Theorem vs. Direct Substitution

Squeeze Theorem Direct Substitution
When to use Function oscillates or is hard to simplify near \( a \) Function is continuous at \( a \)
What you need Two bounding functions with the same limit Just plug \( a \) into \( f(x) \)
Difficulty Finding the right bounds can be tricky Usually straightforward
Classic example \( x^2\sin(1/x) \) as \( x\to 0 \) \( 3x + 2 \) as \( x \to 1 \)

How to Apply the Squeeze Theorem

1
Identify the troublesome factor. Find the part of \( f(x) \) that oscillates or is hard to evaluate (e.g., \( \sin(1/x) \), \( \cos(1/x) \)).
2
Bound the troublesome factor. Use a known inequality like \( -1 \leq \sin(\theta) \leq 1 \) or \( 0 \leq \cos^2(\theta) \leq 1 \).
3
Multiply through by the "nice" factor. Multiply every part of the inequality by the remaining factor, keeping direction if it is non-negative near \( a \).
4
Evaluate the outer limits. Compute \( \lim g(x) \) and \( \lim h(x) \). If they are equal, the Squeeze Theorem gives the answer.

Example 1: \( \lim_{x\to 0} x^2 \sin\!\left(\frac{1}{x}\right) \)

1
Troublesome factor: \( \sin(1/x) \) oscillates wildly near \( x = 0 \).
2
Bound it: We know \( -1 \leq \sin(1/x) \leq 1 \) for all \( x \neq 0 \).
3
Multiply by \( x^2 \) (always \( \geq 0 \)): \( -x^2 \leq x^2\sin(1/x) \leq x^2 \).
4
Evaluate outer limits: \( \lim_{x\to 0}(-x^2)=0 \) and \( \lim_{x\to 0} x^2 = 0 \). Both equal 0.

Example 2: \( \lim_{x\to 0} x\cos\!\left(\frac{1}{x}\right) \)

Step 1: The troublesome factor is \( \cos(1/x) \), which oscillates between \( -1 \) and \( 1 \).

Step 2: Bound it: \( -1 \leq \cos(1/x) \leq 1 \).

Step 3: Near \( x = 0 \), \( x \) can be positive or negative. We use the absolute value trick: $$ -|x| \leq x\cos(1/x) \leq |x|. $$

Step 4: Both \( \lim_{x\to 0}(-|x|) = 0 \) and \( \lim_{x\to 0}|x| = 0 \).

Example 3: The Famous \( \lim_{x\to 0}\frac{\sin x}{x} \)

This classic limit is often proved using a geometric squeeze argument. For \( 0 < x < \frac{\pi}{2} \):

Since \( \lim_{x\to 0}\cos x = 1 \) and \( \lim_{x\to 0} 1 = 1 \), by the Squeeze Theorem:

This result is foundational for deriving the derivatives of \( \sin x \) and \( \cos x \).

Graphing \( \frac{\sin x}{x} \) and Its Bounds

Example 4: \( \lim_{x\to\infty}\frac{\sin x}{x} \)

The Squeeze Theorem also works for limits at infinity. We know:

As \( x \to \infty \), both \( -\frac{1}{x} \to 0 \) and \( \frac{1}{x} \to 0 \).

Even though \( \sin x \) keeps oscillating forever, dividing by \( x \) crushes the oscillation to zero.

Common Mistakes to Avoid

Bounds have different limits
If \( \lim g(x) \neq \lim h(x) \), the Squeeze Theorem cannot conclude anything. You only get a range, not a specific value.
Flipping inequality when multiplying by a negative
If the "nice" factor can be negative near \( a \), you must reverse the inequality signs or use absolute values to maintain valid bounds.
Forgetting to check the inequality near \( a \)
The bounding inequality only needs to hold in a neighborhood of \( a \), not everywhere. But you must verify it holds near \( a \).

Summary of Key Results

Limit Bounds Used Result
\( \lim_{x\to 0} x^2\sin(1/x) \) \( -x^2 \leq f \leq x^2 \) 0
\( \lim_{x\to 0} x\cos(1/x) \) \( -|x| \leq f \leq |x| \) 0
\( \lim_{x\to 0} \frac{\sin x}{x} \) \( \cos x \leq f \leq 1 \) 1
\( \lim_{x\to\infty} \frac{\sin x}{x} \) \( -\frac{1}{x} \leq f \leq \frac{1}{x} \) 0

Question 1 of 5

What is \( \displaystyle\lim_{x\to 0}\, x^2\cos\!\left(\frac{1}{x}\right) \)?

Question 2 of 5

Which conditions are required for the Squeeze Theorem to conclude that \( \lim_{x\to a} f(x) = L \)?

Question 3 of 5

Suppose \( 2x + 1 \leq f(x) \leq x^2 + 2x + 1 \) for all \( x \) near \( 0 \). What is \( \lim_{x\to 0} f(x) \)?

Question 4 of 5

What is \( \displaystyle\lim_{x\to 0}\, \sqrt{x^3 + x^2}\;\sin\!\left(\frac{\pi}{x}\right) \)?

Question 5 of 5

If \( 1 - \frac{x^2}{2} \leq \frac{\sin x}{x} \leq 1 \) near \( x = 0 \), what does the Squeeze Theorem tell us about \( \displaystyle\lim_{x\to 0}\frac{\sin x}{x} \)?