The sandwich theorem and sin θ/θ → 1

The problem

Two friends walk you to the school gate, one on each side, and you stay between them. If both of them reach the gate, where do you end up? At the gate too. You had no choice.

Some limits, like sin θ/θ at θ = 0, give 0/0 and cannot be factorised away. Instead we trap the function between two simpler ones that head to the same number.

Sandwich theorem: if f(x) ≤ g(x) ≤ h(x) near c, and lim(x→c) f(x) = lim(x→c) h(x) = L, then lim(x→c) g(x) = L. Using it, lim(θ→0) sin θ/θ = 1 when θ is in radians.

Key terms

Sandwich theorem

A function trapped between two others that share a limit L must have the limit L too.

Three areas

In a unit circle, triangle OAB fits inside sector OAB, which fits inside triangle OAD. So ½ sin θ < ½ θ < ½ tan θ.

cos θ < sin θ/θ < 1

Divide the area inequality by ½ sin θ and flip each part. As θ → 0, cos θ → 1, so sin θ/θ is squeezed to 1.

Radians only

The sector area ½ r²θ holds only when θ is in radians. In degrees the limit is π/180, not 1.

Short questions with model answers

1. Q1. Evaluate lim(θ→0) sin 7θ/θ.

   • x = 7θ ⇒ θ = x/7; θ → 0 ⇒ x → 0
   • lim(x→0) sin x/(x/7) = 7 · lim(x→0) sin x/x = 7 × 1 = 7

   7

2. Q2. Evaluate lim(θ→0) (1 − cos θ)/θ.

   • (1 − cos θ)/θ × (1 + cos θ)/(1 + cos θ) = sin²θ/(θ(1 + cos θ))
   • (sin θ/θ) · sin θ · 1/(1 + cos θ) → (1)(0)(1/2) = 0

   0

3. Q3. Evaluate lim(x→0) (1 − cos x)/x².

   • 1 − cos x = 2 sin²(x/2)
   • 2 sin²(x/2)/x² = ½ · [sin(x/2)/(x/2)]² → ½ × 1² = ½

   1/2

Common mistakes

• ✗ Writing lim(θ→0) sin 7θ/θ = 1.

  ✓ The result sin u/u → 1 needs the same u on top and bottom. Make the bottom 7θ first: the answer is 7.

• ✗ Using sin x/x → 1 when x is in degrees.

  ✓ The proof uses the sector area ½θ, which needs radians. Since x° = πx/180 radians, lim(x→0) sin x°/x = π/180.

• ✗ Saying lim(x→∞) sin x/x = 1.

  ✓ The result is for x → 0 only. As x → ∞, −1/x ≤ sin x/x ≤ 1/x, and both sides tend to 0, so the sandwich theorem gives 0.

MCQs

1. 1. With θ in radians, lim(θ→0) sin θ/θ equals:

   1. (a) 0
   2. (b) 1
   3. (c) π/180
   4. (d) does not exist
   Show answer

   (b) It is squeezed between cos θ and 1, and cos θ → 1.

2. 2. lim(x→0) sin 7x/x equals:

   1. (a) 1
   2. (b) 1/7
   3. (c) 7
   4. (d) 0
   Show answer

   (c) Write it as 7 · sin 7x/(7x); the fraction tends to 1, leaving 7.

3. 3. lim(x→0) sin ax/sin bx (b ≠ 0) equals:

   1. (a) 1
   2. (b) a/b
   3. (c) b/a
   4. (d) ab
   Show answer

   (b) Write it as (a/b) · [sin ax/(ax)] / [sin bx/(bx)]; both brackets tend to 1.

4. 4. With x in degrees, lim(x→0) sin x°/x equals:

   1. (a) 1
   2. (b) 180/π
   3. (c) π/180
   4. (d) 0
   Show answer

   (c) x° = πx/180 radians, so the limit is (π/180) × 1 = π/180 ≈ 0.01745.

5. 5. lim(x→0) (1 − cos x)/x² equals:

   1. (a) 0
   2. (b) 1
   3. (c) 1/2
   4. (d) 2
   Show answer

   (c) 1 − cos x = 2 sin²(x/2), which gives ½ · [sin(x/2)/(x/2)]² → ½.

Quick revision

• Sandwich theorem: a function trapped between two functions with the same limit L also has limit L.
• From ½ sin θ < ½ θ < ½ tan θ we get cos θ < sin θ/θ < 1, so lim(θ→0) sin θ/θ = 1 in radians.
• Reshape trig limits until sin u/u appears with the same u on top and bottom.