1.8 - Determining Limits Using the Squeeze Theorem

Introduction

The Squeeze Theorem (also called the Sandwich Theorem or Pinching Theorem) is a tool for finding limits of functions that are hard to evaluate directly. The idea is intuitive: if a function is squeezed between two other functions that approach the same limit, the middle function has to approach that limit too.

The Statement

Suppose

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

for all

x

in some interval around

a

(except possibly at

a

itself).

If:

\lim_{x \to a} g(x) = L \quad \text{and} \quad \lim_{x \to a} h(x) = L

Then:

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

In words: if

f(x)

is trapped between

g(x)

and

h(x)

, and both of them approach the same limit

L

x

approaches

a

, then

f(x)

has to approach

L

as well.

Visual Understanding

1.8 - Determining Limits Using the Squeeze Theorem

The function

f(x)

(middle curve) is squeezed between

g(x)

(bottom) and

h(x)

(top). Since both boundary functions approach the same value,

f(x)

has nowhere else to go.

Example 1: Using Trigonometric Bounds

Find

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

Why is this hard? Direct substitution gives

0 \cdot \sin\left(\frac{1}{0}\right)

, which means dividing by zero inside the sine. Undefined.

Step 1: bound the sine function

For any input, sine is bounded:

-1 \leq \sin\left(\frac{1}{x}\right) \leq 1

This is true because sine always oscillates between

-1

and

1

no matter what its argument is.

Step 2: multiply all parts by $x^2$

Since

x^2 \geq 0

for all

x

, multiplying preserves the inequality:

-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2

Step 3: take limits of all three parts

\lim_{x \to 0} (-x^2) \leq \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) \leq \lim_{x \to 0} x^2

Step 4: evaluate the boundary limits

\lim_{x \to 0} (-x^2) = 0 \qquad \lim_{x \to 0} x^2 = 0

Step 5: apply the Squeeze Theorem

Since both boundary limits equal

0

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

Answer:

0

The function

x^2 \sin\left(\frac{1}{x}\right)

oscillates wildly near

x = 0

, but it’s trapped between

-x^2

and

x^2

, both of which approach

0

Example 2: When Limits Don’t Match

Given: -

g(x) = -x^2 + 3x - 5

h(x) = 3x + 1

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

for all

x

Find

\lim_{x \to 1} f(x)

Step 1:

\lim_{x \to 1} g(x) = -(1)^2 + 3(1) - 5 = -1 + 3 - 5 = -3

Step 2:

\lim_{x \to 1} h(x) = 3(1) + 1 = 4

Step 3: since

-3 \neq 4

, the boundary functions approach different values.

The limit cannot be determined with the Squeeze Theorem. All we know is:

-3 \leq \lim_{x \to 1} f(x) \leq 4

f(x)

could approach any value in that range. Without more info about

f(x)

, we can’t pin it down.

Answer: cannot be determined

Example 3: When the Squeeze Theorem Works

Given: -

g(x) = x^3 - 2x^2

h(x) = x^3 + 2x^2

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

for

x

[-2, 2]

Find

\lim_{x \to 0} f(x)

Step 1:

\lim_{x \to 0} g(x) = (0)^3 - 2(0)^2 = 0

Step 2:

\lim_{x \to 0} h(x) = (0)^3 + 2(0)^2 = 0

Step 3: both boundary limits equal

0

, so:

\lim_{x \to 0} f(x) = 0

Justification:

f(x)

is squeezed between two functions that both approach

0

x \to 0

, so by the Squeeze Theorem,

f(x)

must also approach

0

Answer:

0

Checklist for Using the Squeeze Theorem

To use it, you have to verify:

Condition 1: $g(x) \leq f(x) \leq h(x)$ on some interval containing $a$

Condition 2: $\lim_{x \to a} g(x) = L$ (this limit exists)

Condition 3: $\lim_{x \to a} h(x) = L$ (the same value)

Condition 4: both boundary functions are bounded near $a$

If all four hold, then

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

Practice: Can the Squeeze Theorem Be Applied?

Problem 1

$g(x) = -\frac{1}{2}$

$h(x) = \frac{1}{2}$

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

Can we use the Squeeze Theorem to find

\lim_{x \to 0} f(x)

\lim_{x \to 0} g(x) = -\frac{1}{2} \qquad \lim_{x \to 0} h(x) = \frac{1}{2}

Since

-\frac{1}{2} \neq \frac{1}{2}

, the boundary functions approach different values.

No. The Squeeze Theorem doesn’t apply. The limit of

f(x)

could be anything between

-\frac{1}{2}

and

\frac{1}{2}

Problem 2

$g(x) = -x^4$

$h(x) = x^4$

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

Can we find

\lim_{x \to 0} f(x)

\lim_{x \to 0} (-x^4) = 0 \qquad \lim_{x \to 0} x^4 = 0

Both boundary limits equal

0

Yes. The Squeeze Theorem applies, so

\lim_{x \to 0} f(x) = 0

Justification: since

-x^4

and

x^4

both approach

0

x \to 0

, and

f(x)

is squeezed between them,

f(x)

must also approach

0

Problem 3

$g(x) = 1 - x^2$

$h(x) = 1$

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

Can we find

\lim_{x \to 0} f(x)

\lim_{x \to 0} (1 - x^2) = 1 - 0 = 1 \qquad \lim_{x \to 0} 1 = 1

Both boundary limits equal

1

Yes. The Squeeze Theorem applies, so

\lim_{x \to 0} f(x) = 1

Justification:

f(x)

is squeezed between two functions that both approach

1

, so

f(x)

must approach

1

too.

Important: Unbounded Functions

The Squeeze Theorem cannot be applied if either boundary function is unbounded near the point of interest.

Example of unbounded behavior

Consider

g(x) = \frac{1}{x}

x \to 0

: - From the right,

\frac{1}{x} \to +\infty

- From the left,

\frac{1}{x} \to -\infty

That’s unbounded behavior, the function grows without bound.

Why it matters: if one of your boundary functions is unbounded (heads toward

\pm \infty

), the Squeeze Theorem doesn’t apply, because there’s no finite value to squeeze toward.

Example: disqualified by unbounded behavior

$g(x) = \frac{1}{x^2}$

$h(x) = \frac{2}{x^2}$

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

Can we find

\lim_{x \to 0} f(x)

\lim_{x \to 0} \frac{1}{x^2} = +\infty \quad \text{(unbounded)}

\lim_{x \to 0} \frac{2}{x^2} = +\infty \quad \text{(unbounded)}

No. The Squeeze Theorem cannot be applied, because both boundary functions are unbounded. Even though they both “approach” the same thing (infinity), infinity is not a real number, and the theorem requires finite limits.

Common Applications

The Squeeze Theorem is especially useful for:

Products of bounded and vanishing functions, like $x^2 \sin\left(\frac{1}{x}\right)$ as $x \to 0$

Oscillating functions multiplied by functions approaching zero, like $x \cos\left(\frac{1}{x^2}\right)$ as $x \to 0$

Functions with irregular behavior near a point, when direct evaluation is impossible

Proving the special trig limit from 1.6: for $x$ close to $0$ (but not equal to $0$ ), a geometry argument shows $\cos x \leq \frac{\sin x}{x} \leq 1$ . Both bounds approach $1$ as $x \to 0$ , so the Squeeze Theorem gives $\lim_{x \to 0} \frac{\sin x}{x} = 1$ .

Summary: When to Use It

Use the Squeeze Theorem when: - You can’t evaluate the limit directly - The function is bounded between two simpler functions - Both boundary functions approach the same finite limit - Neither boundary function is unbounded

Steps to apply: 1. Identify bounds: find

g(x)

and

h(x)

such that

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

2. Evaluate boundary limits: find

\lim_{x \to a} g(x)

and

\lim_{x \to a} h(x)

3. Check if they’re equal: if both equal

L

, then

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

4. State the conclusion by citing the Squeeze Theorem

Key Takeaways

The Squeeze Theorem is your move when direct evaluation fails

The middle function has to be trapped between two boundary functions

Both boundary functions must approach the same limit

If the boundary limits differ, you cannot determine the middle limit

Unbounded behavior disqualifies the Squeeze Theorem

It shows up a lot with trig functions that oscillate rapidly

For sine and cosine, always remember $-1 \leq \sin(\text{anything}) \leq 1$

Multiply inequalities by the same function to create your bounds

Always justify by showing both boundary limits equal the same value

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