Mastering Trig Limits with the Squeeze Theorem

A student stares at lim_x→0 sin x / x and assumes the answer is 0 since sin 0 = 0. That mistake costs exam points. The actual limit is 1, and it is not obvious. The reason involves geometry, not algebra, and it makes trigonometric limits a separate skill from polynomial limits. The two special limits that govern trig limits are lim_x→0 sin x / x = 1 and lim_x→0 (cos x, 1) / x = 0. Stewart §3.3 gives both formulas. The first one is the foundation; the second is derived from it using the identity cos x, 1 =, 2 sin²(x/2). Once you know these two, every other limit involving sin, cos and tan reduces to them.

The Squeeze Theorem makes the sin x / x limit rigorous without epsilon-delta. Draw a unit circle and a sector of angle x (in radian measure). Three areas appear: the triangle inside the sector, the sector itself, and the triangle that contains the sector. For x between 0 and π/2, the inequalities sin x ≤ x ≤ tan x hold. Divide by sin x to get 1 ≤ x / sin x ≤ 1 / cos x. Take reciprocals, which reverses the inequalities: cos x ≤ sin x / x ≤ 1. As x → 0, cos x → 1. The Squeeze Theorem forces sin x / x → 1. The same argument works from the left because both sin x and x are odd functions. This proof is the standard one from Stewart §2.3 and §3.3. The Squeeze Theorem applies here as sin x / x is bounded between two functions that share the same limit.

When the argument inside the sine is a multiple of x, rescale. Compute lim_x→0 sin(3x) / x. Multiply numerator and denominator by 3: (3 sin(3x)) / (3x) = 3 · sin(3x) / (3x). As x → 0, 3x → 0, so sin(3x) / (3x) → 1. The limit is 3. The general rule is lim_x→0 sin(kx) / x = k. The same method works for tan(kx) / x. Write tan(kx) = sin(kx) / cos(kx). The cos term goes to 1, so the limit becomes k / 1 = k. This rescaling technique appears in every trig limits example set. If the denominator is not x but something like 2x, factor it: lim_x→0 sin(5x) / (2x) = (5/2) · sin(5x) / (5x) → 5/2.

The limit lim_x→0 x · sin(1 / x) is a classic Squeeze Theorem application. As x approaches 0, 1/x grows without bound, and sin(1/x) oscillates between -1 and 1. Direct substitution gives 0 · oscillating, which is an indeterminate form 0 · (bounded). Because |sin(1/x)| ≤ 1, multiply by |x| to get, -|x| ≤ x · sin(1/x) ≤ |x|. As x → 0, both, -|x| and |x| approach 0. By the Squeeze Theorem, the limit is 0. This example contrasts with lim_x→0 sin(1/x), which does not exist because the oscillation never settles. The factor x forces the product to zero. This is a common exam trick and appears in Stewart §2.3 exercises.

The formula lim_x→0 sin x / x = 1 is true only when x is measured in radian measure. In degrees, sin x is periodic with period 360, but the limit changes. Convert a degree input to radian measure first. If the problem says lim_x→0 sin(x°) / x, rewrite x° as πx / 180 radians. The limit becomes lim_x→0 sin(πx / 180) / x = π/180. The Squeeze Theorem proof uses arc length on a unit circle, which is defined in radian measure. Using degrees breaks the geometric inequality sin x ≤ x ≤ tan x because the arc length and the angle measure no longer match. Every calculus textbook, including Stewart §3.3, assumes radian measure for derivative formulas. If the exam question does not specify, assume radian measure.

Example 1: lim_x→0 tan(2x) / (5x)

Write tan(2x) = sin(2x) / cos(2x). The limit becomes lim_x→0 [sin(2x) / (5x)] · [1 / cos(2x)]. Rescale the sine term: sin(2x) / (5x) = (2/5) · sin(2x) / (2x) → 2/5. The cos term goes to 1. The answer is 2/5.

Example 2: lim_x→0 (1, cos x) / x²

Use the identity 1, cos x = 2 sin²(x/2). The limit becomes 2 · lim_x→0 sin²(x/2) / x². Write as 2 · [sin(x/2) / x]². Rescale: sin(x/2) / x = (1/2) · sin(x/2) / (x/2). The square gives (1/2)² = 1/4. Multiply by the 2 outside: 2 · 1/4 = 1/2. Stewart §3.3 lists this result.

Example 3: lim_x→0 sin(4x) / sin(6x)

Divide numerator and denominator by x: [sin(4x) / x] / [sin(6x) / x]. Each term rescales: sin(4x) / x → 4, sin(6x) / x → 6. The limit is 4/6 = 2/3.

Example 4: lim_x→0 (x + sin x) / x

Split the fraction: x/x + sin x / x = 1 + sin x / x → 1 + 1 = 2. This uses the sum law from Stewart §2.3.

Trig limits appear in every AP Calculus AB exam on the limits unit. They reappear when computing derivatives of sine and cosine, because the derivative formula uses the difference quotient that produces sin(Δx) / Δx. They also appear in polar curve slopes in BC calculus. L'Hôpital's rule can solve these limits too, but it requires derivatives that themselves rely on these limits, creating a circular argument if used in a proof. The Squeeze Theorem is the honest method. Factoring and rationalizing are not relevant here; trig limits are their own category. If you rely solely on L'Hôpital's rule, you lose the geometric intuition that the Squeeze Theorem provides.

Students apply the Squeeze Theorem to limits that are not bounded or pick bounding functions that do not share the same limit. For lim_x→0 sin(1/x), bounding between, 1 and 1 is correct, but the bounds have different limits (1 and, 1), so the theorem does not apply. The limit does not exist. The Squeeze Theorem requires both bounds to converge to the same number. Check that condition before writing the answer.