Calculus-Master

Trig Limits & the Squeeze Theorem: Step-by-Step Guide

Some limits can't be found by plugging in or factoring. Trig limits like lim⁡x→0sin⁡5xx\lim_{x \to 0} \frac{\sin 5x}{x} are solved by rewriting them into the special limit sin⁡θθ→1\frac{\sin \theta}{\theta} \to 1, and oscillating limits like lim⁡x→0x2sin⁡(1x)\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) are solved by trapping the function between two simpler ones with the Squeeze Theorem.

Core Theorem
Squeeze Theorem: if g(x)≤f(x)≤h(x)g(x) \le f(x) \le h(x) near x=cx = c and
lim⁡x→cg(x)=lim⁡x→ch(x)=L,\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L,
then lim⁡x→cf(x)=L\lim_{x \to c} f(x) = L.
Special trig limits (xx in radians):
lim⁡x→0sin⁡xx=1,lim⁡x→01−cos⁡xx=0\lim_{x \to 0} \frac{\sin x}{x} = 1, \qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0
Step-by-Step SOP
  1. 1

    1. Try direct substitution

    If plugging in gives a real number, that's the limit. Only an indeterminate form like 00\frac{0}{0} needs more work.
  2. 2

    2. Trig 0/0: match the angle

    Rewrite so each sin⁡(kx)\sin(kx) is divided by exactly kxkx: sin⁡5xx=5⋅sin⁡5x5x\frac{\sin 5x}{x} = 5 \cdot \frac{\sin 5x}{5x}. Then each matched piece goes to 11.
  3. 3

    3. Oscillating factor: bound it

    Start from −1≤sin⁡(anything)≤1-1 \le \sin(\text{anything}) \le 1 (same for cosine) and multiply through by the other factor. If that factor is never negative (like x2x^2), the inequality keeps its direction.
  4. 4

    4. Check both bounds agree

    Find the limits of the lower and upper functions. The Squeeze Theorem only works if they are equal.
  5. 5

    5. Write the justification

    On the AP exam, state the inequality, the two equal limits, and name the theorem: "Since g(x)≤f(x)≤h(x)g(x) \le f(x) \le h(x) and both gg and hh approach LL, by the Squeeze Theorem lim⁡f(x)=L\lim f(x) = L."

Practice Exercises

Example 01Easy
Evaluate lim⁡x→0x2sin⁡(1x)\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right).
NEED A HINT?
sin⁡(1x)\sin\left(\frac{1}{x}\right) oscillates forever as x→0x \to 0, but it always stays between −1-1 and 11.
SHOW DETAILED EXPLANATION

Step 1: Bound the oscillating part

−1≤sin⁡(1x)≤1-1 \le \sin\left(\frac{1}{x}\right) \le 1 for all x≠0x \ne 0.

Step 2: Multiply by x²

Since x2≥0x^2 \ge 0, the inequality keeps its direction: −x2≤x2sin⁡(1x)≤x2-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2.

Step 3: Limits of the bounds

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

Step 4: Squeeze

Both bounds approach 00, so by the Squeeze Theorem lim⁡x→0x2sin⁡(1x)=0\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0.
Example 02Easy
Evaluate lim⁡θ→0sin⁡2θ3θ\lim_{\theta \to 0} \frac{\sin 2\theta}{3\theta}.
NEED A HINT?
The angle is 2θ2\theta, so you want 2θ2\theta in the denominator.
SHOW DETAILED EXPLANATION

Step 1: Pull out the constant

sin⁡2θ3θ=13⋅sin⁡2θθ\frac{\sin 2\theta}{3\theta} = \frac{1}{3} \cdot \frac{\sin 2\theta}{\theta}.

Step 2: Match the angle

sin⁡2θθ=2⋅sin⁡2θ2θ\frac{\sin 2\theta}{\theta} = 2 \cdot \frac{\sin 2\theta}{2\theta}.

Step 3: Use the special limit

As θ→0\theta \to 0, 2θ→02\theta \to 0, so sin⁡2θ2θ→1\frac{\sin 2\theta}{2\theta} \to 1. The limit is 13⋅2⋅1=23\frac{1}{3} \cdot 2 \cdot 1 = \frac{2}{3}.
Example 03Medium
Evaluate lim⁡x→0sin⁡3xsin⁡7x\lim_{x \to 0} \frac{\sin 3x}{\sin 7x}.
NEED A HINT?
Divide the top and bottom by xx, then match each angle separately.
SHOW DETAILED EXPLANATION

Step 1: Rewrite with matched angles

sin⁡3xsin⁡7x=3⋅sin⁡3x3x7⋅sin⁡7x7x\frac{\sin 3x}{\sin 7x} = \frac{3 \cdot \frac{\sin 3x}{3x}}{7 \cdot \frac{\sin 7x}{7x}}.

Step 2: Take the limit

Both fractions go to 11, so the limit is 37\frac{3}{7}.
Example 04Hard
Evaluate lim⁡x→01−cos⁡xx2\lim_{x \to 0} \frac{1 - \cos x}{x^2}.
NEED A HINT?
Multiply the top and bottom by the conjugate 1+cos⁡x1 + \cos x and use 1−cos⁡2x=sin⁡2x1 - \cos^2 x = \sin^2 x.
SHOW DETAILED EXPLANATION

Step 1: Multiply by the conjugate

1−cos⁡xx2⋅1+cos⁡x1+cos⁡x=sin⁡2xx2(1+cos⁡x)\frac{1 - \cos x}{x^2} \cdot \frac{1 + \cos x}{1 + \cos x} = \frac{\sin^2 x}{x^2(1 + \cos x)}.

Step 2: Split into known limits

=(sin⁡xx)2⋅11+cos⁡x= \left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1 + \cos x}.

Step 3: Take the limit

12⋅11+1=121^2 \cdot \frac{1}{1 + 1} = \frac{1}{2}.
Example 05AB/BC Standard
AP style. Suppose 2x+1≤f(x)≤x2+22x + 1 \le f(x) \le x^2 + 2 for all xx. Find lim⁡x→1f(x)\lim_{x \to 1} f(x) and justify your answer.
NEED A HINT?
You don't need a formula for ff. Find the limits of the two bounding functions at x=1x = 1.
SHOW DETAILED EXPLANATION

Step 1: Lower bound

lim⁡x→1(2x+1)=3\lim_{x \to 1} (2x + 1) = 3.

Step 2: Upper bound

lim⁡x→1(x2+2)=3\lim_{x \to 1} (x^2 + 2) = 3.

Step 3: Justify with the Squeeze Theorem

Since 2x+1≤f(x)≤x2+22x + 1 \le f(x) \le x^2 + 2 and both bounds approach 33 as x→1x \to 1, by the Squeeze Theorem lim⁡x→1f(x)=3\lim_{x \to 1} f(x) = 3. (The bounds are consistent: x2+2−(2x+1)=(x−1)2≥0x^2 + 2 - (2x + 1) = (x - 1)^2 \ge 0.)
Example 06Medium
Evaluate lim⁡x→∞sin⁡xx\lim_{x \to \infty} \frac{\sin x}{x}.
NEED A HINT?
This is NOT the special limit: x→∞x \to \infty, not 00. Bound sin⁡x\sin x instead.
SHOW DETAILED EXPLANATION

Step 1: Bound the numerator

For x>0x > 0: −1x≤sin⁡xx≤1x-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}.

Step 2: Squeeze

Both −1x-\frac{1}{x} and 1x\frac{1}{x} approach 00 as x→∞x \to \infty, so lim⁡x→∞sin⁡xx=0\lim_{x \to \infty} \frac{\sin x}{x} = 0.
Example 07Easy
Practice: (a) lim⁡x→0x4cos⁡(2x)\lim_{x \to 0} x^4 \cos\left(\frac{2}{x}\right) (b) lim⁡θ→0sin⁡5θ4θ\lim_{\theta \to 0} \frac{\sin 5\theta}{4\theta}
NEED A HINT?
(a) Bound cos⁡(2x)\cos\left(\frac{2}{x}\right) between −1-1 and 11. (b) Match the angle 5θ5\theta.
SHOW DETAILED EXPLANATION

(a) Squeeze

−x4≤x4cos⁡(2x)≤x4-x^4 \le x^4 \cos\left(\frac{2}{x}\right) \le x^4 and both bounds go to 00, so the limit is 00.

(b) Match the angle

sin⁡5θ4θ=54⋅sin⁡5θ5θ→54\frac{\sin 5\theta}{4\theta} = \frac{5}{4} \cdot \frac{\sin 5\theta}{5\theta} \to \frac{5}{4}.
Common Pitfalls
  • ⚠
    Angle doesn't matchlim⁡x→0sin⁡5xx\lim_{x \to 0} \frac{\sin 5x}{x} is 55, not 11. The special limit only applies when the denominator is exactly the angle inside the sine.
  • ⚠
    Using sin x / x → 1 at infinitysin⁡xx→1\frac{\sin x}{x} \to 1 only as x→0x \to 0. As x→∞x \to \infty it goes to 00 (squeeze it).
  • ⚠
    Bounds with different limitsIf the lower and upper functions approach different values, the Squeeze Theorem tells you nothing. Find tighter bounds or another method.
  • ⚠
    Multiplying by a negative factorMultiplying −1≤sin⁡(⋅)≤1-1 \le \sin(\cdot) \le 1 by a factor that can be negative flips the inequality. Use ∣f(x)∣≤|f(x)| \le (something) or pick a factor like x2x^2 that is never negative.
  • ⚠
    Degrees instead of radianssin⁡xx→1\frac{\sin x}{x} \to 1 is only true with xx in radians. AP Calculus always uses radians.
Frequently Asked Questions

What is the Squeeze Theorem in simple terms?

If a function is trapped between two other functions, and those two approach the same value at a point, the trapped function must approach that value too. It is also called the Sandwich Theorem or Pinching Theorem.

When should you use the Squeeze Theorem?

When a limit has a bounded, oscillating factor (like sin⁡(1x)\sin\left(\frac{1}{x}\right) or cos⁡(2x)\cos\left(\frac{2}{x}\right)) multiplied by something that goes to 00, or when a problem gives you inequalities around f(x)f(x) instead of a formula.

How do you solve trig limits when the Squeeze Theorem doesn't work?

Rewrite the limit into the special forms sin⁡θθ→1\frac{\sin \theta}{\theta} \to 1 or 1−cos⁡θθ→0\frac{1 - \cos \theta}{\theta} \to 0 by matching angles, multiplying by a conjugate, or using identities like tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}. Later in the course (Unit 4), L'Hôpital's Rule also handles 00\frac{0}{0} trig limits.

Why does sin(x)/x approach 1?

Geometrically, for small angles (in radians) the arc length xx, the height sin⁡x\sin x and the tangent length tan⁡x\tan x are nearly equal. Comparing the areas of a triangle, a sector and a larger triangle gives cos⁡x≤sin⁡xx≤1\cos x \le \frac{\sin x}{x} \le 1, and the Squeeze Theorem finishes the proof.

Is the Squeeze Theorem on the AP Calculus AB exam?

Yes. It is part of Unit 1 (Limits and Continuity). Expect to apply it in multiple choice and to name it as justification when a free-response question gives you inequalities around a function.

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