Day 3

Squeeze Theorem

The Squeeze Theorem Special Trigonometric Limits

The Squeeze Theorem

For functions that oscillate wildly (like $\sin(1/x)$ ), direct substitution and algebra don't work — instead we trap the function between two simpler ones that share the same limit.

Core Theorem

g(x) \le f(x) \le h(x)

near

x=c

, and

\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L

, then

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

Watch the TikTok ExplanationThe Squeeze Theorem→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Evaluate

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

How to think about it

Reach for this when the function itself is hard but you can trap it between two easy functions that share the same limit. Start from a bound you are certain of (

-1 \le \sin(\cdot) \le 1

) and build the inequality outward.

Key idea: Start from the bounded range of $\sin(\cdot)$ , then multiply through by $x^2$ .

Full solution

1. Step 1: Start with the Bounded Range

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

for all

x \neq 0

2. Step 2: Multiply by $x^2$

Since

x^2 \ge 0

, the inequality direction stays the same:

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

3. Step 3: Evaluate the Outer Limits

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

and

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

4. Step 4: Apply the Squeeze Theorem

Since both outer limits equal 0,

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

Step-by-Step SOP

1
Find a Bounded Piece
Identify the oscillating factor (usually $\sin$ or $\cos$ of something) and write its natural $[-1, 1]$ bound.
2
Multiply Through Carefully
Multiply the inequality by the remaining factor, watching whether it's non-negative.
3
Evaluate Both Outer Limits
If both outer bounds converge to the same value $L$ , the squeezed function also converges to $L$ .

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Evaluate

\lim_{x \to 0} \left[ x^4 \cos\left(\frac{5}{x}\right) \right]

Need a hint?

Same technique as Example 1 — start with the range of

\cos(\cdot)

Show solution

1. Step 1: Bound the Cosine Term

-1 \le \cos\left(\frac{5}{x}\right) \le 1

for all

x \neq 0

2. Step 2: Multiply by $x^4$

-x^4 \le x^4 \cos\left(\frac{5}{x}\right) \le x^4

, since

x^4 \ge 0

3. Step 3: Squeeze

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

, so

\lim_{x \to 0} x^4 \cos\left(\frac{5}{x}\right) = 0

Practice 02Medium

1 - \frac{x^2}{4} \le u(x) \le 1 + \frac{x^2}{2}

for all

x \neq 0

, find

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

Need a hint?

You're already given both bounding functions directly — you don't need to build them yourself this time.

Show solution

1. Step 1: Evaluate the Lower Bound's Limit

\lim_{x \to 0} \left(1 - \frac{x^2}{4}\right) = 1 - 0 = 1

2. Step 2: Evaluate the Upper Bound's Limit

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

3. Step 3: Apply the Squeeze Theorem

Since both bounds converge to 1,

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

— even without knowing a formula for

u(x)

itself.

Common Pitfalls

⚠
You Cannot Just 'Plug In' $\sin(1/x)$ and $\cos(1/x)$ have no limit at all as $x \to 0$ (they oscillate infinitely) — the trick only works because you multiply by a term ( $x^2$ , $x^4$ ) that squeezes both bounds to 0.
⚠
Check the Inequality DirectionMultiplying an inequality by a negative quantity flips it. Always confirm the multiplier ( $x^2$ , $x^4$ , etc.) is non-negative before keeping the same direction.

Special Trigonometric Limits

Two trigonometric limit identities show up constantly on the AP exam — memorize them and learn to spot the pattern even when the angle is disguised.

Core Theorem

\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1

\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = 0

These hold whenever the expression inside the trig function matches the denominator exactly.

Watch the TikTok ExplanationLimits of Trigonometric Functions→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Evaluate

\lim_{\theta \to 0} \frac{\sin 2\theta}{3\theta}

How to think about it

\frac{\sin(\text{angle})}{\text{that same angle}} \to 1

only when the two angles match exactly. When they don't (here

2\theta

3\theta

), multiply by a form of

1

to force a match, and carry the leftover constant out front.

Key idea: The angle inside $\sin$ is $2\theta$ , but the denominator is $3\theta$ — force them to match by multiplying by a clever form of 1.

Full solution

1. Step 1: Pull Out the Constant

\frac{1}{3} \cdot \lim_{\theta \to 0} \frac{\sin 2\theta}{\theta}

2. Step 2: Match the Angle to the Denominator

Multiply numerator and denominator by 2:

\frac{1}{3} \cdot \lim_{\theta \to 0} \left[ \frac{\sin 2\theta}{2\theta} \cdot 2 \right]

3. Step 3: Apply the Identity

2\theta \to 0

\frac{\sin 2\theta}{2\theta} \to 1

, leaving

\frac{2}{3} \cdot 1

4. Step 4: Final Answer

\frac{2}{3}

Step-by-Step SOP

1
Identify the Mismatch
Compare the angle inside the trig function to the denominator.
2
Multiply by a Clever Form of 1
Scale the numerator and denominator so the angle matches, then pull the resulting constant outside the limit.
3
Apply the Identity
Once the angle matches the denominator exactly, substitute the known limit (1 or 0).

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Easy

Evaluate

\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta}

Need a hint?

This is exactly the second special trig limit — no algebra needed.

Show solution

1. Step 1: Recognize the Identity

This expression exactly matches

\lim_{\theta \to 0} \frac{1-\cos\theta}{\theta}

2. Step 2: Apply It Directly

By the special trig limit, this equals

0

Practice 02Hard

Evaluate

\lim_{x \to 0} x \cot x

Need a hint?

Rewrite

\cot x

\frac{\cos x}{\sin x}

, then regroup so both special trig limits appear.

Show solution

1. Step 1: Rewrite Cotangent

x \cot x = x \cdot \frac{\cos x}{\sin x} = \frac{x}{\sin x} \cdot \cos x

2. Step 2: Split Into Known Pieces

\frac{x}{\sin x} = \frac{1}{\sin x / x}

, so the whole expression is

\frac{1}{\frac{\sin x}{x}} \cdot \cos x

3. Step 3: Apply Both Special Limits

x \to 0

\frac{\sin x}{x} \to 1

and

\cos x \to 1

4. Step 4: Final Answer

\frac{1}{1} \cdot 1 = 1

Common Pitfalls

⚠
The Angle Must Match the Denominator $\lim_{\theta \to 0} \frac{\sin 2\theta}{\theta} \neq 1$ directly — you must algebraically force the denominator to also read $2\theta$ before applying the identity.
⚠
θ Must Approach 0These identities only apply as the angle approaches 0, not as it approaches any other value.

Gary Chang

Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

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Worked Examples

1. Step 1: Start with the Bounded Range

2. Step 2: Multiply by $x^2$

3. Step 3: Evaluate the Outer Limits

4. Step 4: Apply the Squeeze Theorem

Find a Bounded Piece

Multiply Through Carefully

Evaluate Both Outer Limits

Practice — Your Turn

1. Step 1: Bound the Cosine Term

2. Step 2: Multiply by $x^4$

3. Step 3: Squeeze

1. Step 1: Evaluate the Lower Bound's Limit

2. Step 2: Evaluate the Upper Bound's Limit

3. Step 3: Apply the Squeeze Theorem

Worked Examples

1. Step 1: Pull Out the Constant

2. Step 2: Match the Angle to the Denominator

3. Step 3: Apply the Identity

4. Step 4: Final Answer

Identify the Mismatch

Multiply by a Clever Form of 1

Apply the Identity

Practice — Your Turn

1. Step 1: Recognize the Identity

2. Step 2: Apply It Directly

1. Step 1: Rewrite Cotangent

2. Step 2: Split Into Known Pieces

3. Step 3: Apply Both Special Limits

4. Step 4: Final Answer

Gary Chang