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Day 3

Squeeze Theorem


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
If g(x)f(x)h(x)g(x) \le f(x) \le h(x) near x=cx=c, and limxcg(x)=limxch(x)=L\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L, then limxcf(x)=L\lim_{x \to c} f(x) = L.
Watch the TikTok ExplanationThe Squeeze Theorem
Step-by-Step SOP
  1. 1

    Find a Bounded Piece

    Identify the oscillating factor (usually sin\sin or cos\cos of something) and write its natural [1,1][-1, 1] bound.
  2. 2

    Multiply Through Carefully

    Multiply the inequality by the remaining factor, watching whether it's non-negative.
  3. 3

    Evaluate Both Outer Limits

    If both outer bounds converge to the same value LL, the squeezed function also converges to LL.

Practice Exercises


Example 01Medium
Evaluate limx0[x2sin(1x)]\lim_{x \to 0} \left[ x^2 \sin\left(\frac{1}{x}\right) \right].
Watch the TikTok ExplanationThe Squeeze Theorem
NEED A HINT?
Start from the bounded range of sin()\sin(\cdot), then multiply through by x2x^2.
SHOW DETAILED EXPLANATION

Step 1: Start with the Bounded Range

1sin(1x)1-1 \le \sin\left(\frac{1}{x}\right) \le 1 for all x0x \neq 0.

Step 2: Multiply by $x^2$

Since x20x^2 \ge 0, the inequality direction stays the same: x2x2sin(1x)x2-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2.

Step 3: Evaluate the Outer Limits

limx0(x2)=0\lim_{x \to 0} (-x^2) = 0 and limx0(x2)=0\lim_{x \to 0} (x^2) = 0.

Step 4: Apply the Squeeze Theorem

Since both outer limits equal 0, limx0x2sin(1x)=0\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0.
Example 02Medium
Evaluate limx0[x4cos(5x)]\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()\cos(\cdot).
SHOW DETAILED EXPLANATION

Step 1: Bound the Cosine Term

1cos(5x)1-1 \le \cos\left(\frac{5}{x}\right) \le 1 for all x0x \neq 0.

Step 2: Multiply by $x^4$

x4x4cos(5x)x4-x^4 \le x^4 \cos\left(\frac{5}{x}\right) \le x^4, since x40x^4 \ge 0.

Step 3: Squeeze

limx0(±x4)=0\lim_{x \to 0} (\pm x^4) = 0, so limx0x4cos(5x)=0\lim_{x \to 0} x^4 \cos\left(\frac{5}{x}\right) = 0.
Example 03Medium
If 1x24u(x)1+x221 - \frac{x^2}{4} \le u(x) \le 1 + \frac{x^2}{2} for all x0x \neq 0, find limx0u(x)\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 DETAILED EXPLANATION

Step 1: Evaluate the Lower Bound's Limit

limx0(1x24)=10=1\lim_{x \to 0} \left(1 - \frac{x^2}{4}\right) = 1 - 0 = 1.

Step 2: Evaluate the Upper Bound's Limit

limx0(1+x22)=1+0=1\lim_{x \to 0} \left(1 + \frac{x^2}{2}\right) = 1 + 0 = 1.

Step 3: Apply the Squeeze Theorem

Since both bounds converge to 1, limx0u(x)=1\lim_{x \to 0} u(x) = 1 — even without knowing a formula for u(x)u(x) itself.
Common Pitfalls
  • You Cannot Just 'Plug In'sin(1/x)\sin(1/x) and cos(1/x)\cos(1/x) have no limit at all as x0x \to 0 (they oscillate infinitely) — the trick only works because you multiply by a term (x2x^2, x4x^4) that squeezes both bounds to 0.
  • Check the Inequality DirectionMultiplying an inequality by a negative quantity flips it. Always confirm the multiplier (x2x^2, x4x^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
limθ0sinθθ=1\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1
limθ01cosθθ=0\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
Step-by-Step SOP
  1. 1

    Identify the Mismatch

    Compare the angle inside the trig function to the denominator.
  2. 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. 3

    Apply the Identity

    Once the angle matches the denominator exactly, substitute the known limit (1 or 0).

Practice Exercises


Example 01Medium
Evaluate limθ0sin2θ3θ\lim_{\theta \to 0} \frac{\sin 2\theta}{3\theta}.
Watch the TikTok ExplanationLimits of Trigonometric Functions
NEED A HINT?
The angle inside sin\sin is 2θ2\theta, but the denominator is 3θ3\theta — force them to match by multiplying by a clever form of 1.
SHOW DETAILED EXPLANATION

Step 1: Pull Out the Constant

13limθ0sin2θθ\frac{1}{3} \cdot \lim_{\theta \to 0} \frac{\sin 2\theta}{\theta}.

Step 2: Match the Angle to the Denominator

Multiply numerator and denominator by 2: 13limθ0[sin2θ2θ2]\frac{1}{3} \cdot \lim_{\theta \to 0} \left[ \frac{\sin 2\theta}{2\theta} \cdot 2 \right].

Step 3: Apply the Identity

As 2θ02\theta \to 0, sin2θ2θ1\frac{\sin 2\theta}{2\theta} \to 1, leaving 231\frac{2}{3} \cdot 1.

Step 4: Final Answer

23\frac{2}{3}.
Example 02Easy
Evaluate limθ01cosθθ\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta}.
NEED A HINT?
This is exactly the second special trig limit — no algebra needed.
SHOW DETAILED EXPLANATION

Step 1: Recognize the Identity

This expression exactly matches limθ01cosθθ\lim_{\theta \to 0} \frac{1-\cos\theta}{\theta}.

Step 2: Apply It Directly

By the special trig limit, this equals 00.
Example 03Hard
Evaluate limx0xcotx\lim_{x \to 0} x \cot x.
NEED A HINT?
Rewrite cotx\cot x as cosxsinx\frac{\cos x}{\sin x}, then regroup so both special trig limits appear.
SHOW DETAILED EXPLANATION

Step 1: Rewrite Cotangent

xcotx=xcosxsinx=xsinxcosxx \cot x = x \cdot \frac{\cos x}{\sin x} = \frac{x}{\sin x} \cdot \cos x.

Step 2: Split Into Known Pieces

xsinx=1sinx/x\frac{x}{\sin x} = \frac{1}{\sin x / x}, so the whole expression is 1sinxxcosx\frac{1}{\frac{\sin x}{x}} \cdot \cos x.

Step 3: Apply Both Special Limits

As x0x \to 0: sinxx1\frac{\sin x}{x} \to 1 and cosx1\cos x \to 1.

Step 4: Final Answer

111=1\frac{1}{1} \cdot 1 = 1.
Common Pitfalls
  • The Angle Must Match the Denominatorlimθ0sin2θθ1\lim_{\theta \to 0} \frac{\sin 2\theta}{\theta} \neq 1 directly — you must algebraically force the denominator to also read 2θ2\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

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.