What is the limit of sin θ/θ as θ approaches 0?

When θ is in radians, lim(θ→0) sin θ/θ = 1. It is proved by squeezing sin θ/θ between cos θ and 1.

Why must θ be in radians?

The proof uses the sector area ½r²θ, which is true only in radians. In degrees, lim(x→0) sin x°/x = π/180.

What is the sandwich theorem?

If f(x) ≤ g(x) ≤ h(x) near c and f and h have the same limit L at c, then g also has the limit L at c.

What is lim(θ→0) sin 7θ/θ?

Write it as 7 · sin 7θ/(7θ). The fraction tends to 1, so the limit is 7.

What is lim(x→0) (1 − cos x)/x²?

Using 1 − cos x = 2 sin²(x/2), the expression becomes ½ · [sin(x/2)/(x/2)]², which tends to ½.

Limit of sin θ/θ as θ → 0: Proof and Solved Examples (2nd Year Maths)

When θ is measured in radians, lim(θ→0) sin θ/θ = 1. The proof traps sin θ/θ between cos θ and 1 using three areas in a unit circle, and the sandwich theorem does the rest. In degrees the limit is π/180, not 1.

Direct substitution gives 0/0, which tells us nothing, and no factorising removes it. So the textbook uses a smarter idea: squeeze the function between two others that head to the same number.

The sandwich theorem

If f(x) ≤ g(x) ≤ h(x) near c, and both f(x) and h(x) tend to the same limit L as x → c, then g(x) also tends to L. It is like walking to the college gate with a friend on each side: if both reach the gate, so do you.

The proof in four lines

Take a unit circle (radius 1) and a small positive angle θ in radians. Compare three areas:

Area of triangle OAB = ½ sin θ
Area of sector OAB = ½ θ (this needs radians)
Area of triangle OAD = ½ tan θ

The triangle fits inside the sector, and the sector inside the bigger triangle, so ½ sin θ < ½ θ < ½ tan θ. Divide by ½ sin θ and take reciprocals: cos θ < sin θ/θ < 1. As θ → 0, cos θ → 1, so sin θ/θ is squeezed to 1.

You can see the squeeze in numbers: at θ = 0.1, cos θ = 0.995004 and sin θ/θ = 0.998334, trapped just below 1. At θ = 1 the gap 1 − cos θ is 0.459698; at θ = 0.01 it is only 0.000050.

Solved examples

Example 1: lim(θ→0) sin 7θ/θ Put x = 7θ, so θ = x/7 and x → 0. Then sin x/(x/7) = 7 · sin x/x → 7 × 1 = 7.

Example 2: lim(θ→0) (1 − cos θ)/θ Multiply by (1 + cos θ)/(1 + cos θ): the top becomes 1 − cos²θ = sin²θ. So the limit is (sin θ/θ) · sin θ · 1/(1 + cos θ) → (1)(0)(½) = 0.

Example 3: lim(x→0) (1 − cos x)/x² Use 1 − cos x = 2 sin²(x/2). Then 2 sin²(x/2)/x² = ½ · [sin(x/2)/(x/2)]² → ½ × 1² = ½.

Example 4: lim(x→0) sin ax/sin bx Write it as (a/b) · [sin ax/(ax)] ÷ [sin bx/(bx)]. Both brackets tend to 1, so the limit is a/b.

Common mistakes in the exam

Writing sin 7θ/θ → 1. The result sin u/u → 1 needs the same u on top and bottom. Make the bottom 7θ first.
Using it in degrees. The sector area ½θ needs radians. Since x° = πx/180 radians, lim(x→0) sin x°/x = π/180 ≈ 0.01745.
Using it as x → ∞. There, −1/x ≤ sin x/x ≤ 1/x and both sides tend to 0, so the limit is 0, not 1.

Quick revision

lim(θ→0) sin θ/θ = 1, θ in radians.
Proof: ½ sin θ < ½ θ < ½ tan θ ⇒ cos θ < sin θ/θ < 1.
Reshape every trig limit until sin u/u appears with the same u.

Try the unit-circle squeeze yourself, with more solved limits and MCQs, in the free lesson The sandwich theorem and sin θ/θ.