Single-Variable-Calculus · Unit 3 · Video 3 · Interactive Practice

The Two Trig Limits, Proved at Last | The Bow and the Bowstring

IKey Formulas

StatementNameWhat it says about the picture
lim⁡θ→0sin⁡θθ=1\lim\limits_{\theta \to 0} \dfrac{\sin\theta}{\theta} = 1Limit Bbowstring ÷\div bow →1\to 1
lim⁡θ→0cos⁡θ−1θ=0\lim\limits_{\theta \to 0} \dfrac{\cos\theta - 1}{\theta} = 0Limit Agap ÷\div arc →0\to 0
bowstring=2sin⁡θ,bow=2θ\text{bowstring} = 2\sin\theta, \quad \text{bow} = 2\thetaThe doubled picture2sin⁡θ2θ=sin⁡θθ\dfrac{2\sin\theta}{2\theta} = \dfrac{\sin\theta}{\theta}
gap=1−cos⁡θ\text{gap} = 1 - \cos\thetaLimit A's numeratorthe sliver from the bowstring out to the bow

Key Insight: Short pieces of curves are nearly straight. And θ\theta can be a length on the board at all only because it is measured in radians: the angle at the centre and the arc it cuts off are then the same number.

IIVisualization 1 — Bow and Bowstring

Bowstring and bow share both endpoints and differ only in the route between them.

💡 The principle is not about circles: any smooth curve, cut short enough, is nearly the straight segment joining its ends.

IIIVisualization 2 — Where 1−cos⁡θ1 - \cos\theta Lives

Every term of limit A is already a length in this one picture.

Step 1 — The centre line is a radius
vertex⟶circle  =  1\text{vertex} \longrightarrow \text{circle} \;=\; 1

IVVisualization 3 — Send the Vertex Away

The gap closes to nothing while the arc keeps its length.

💡 Since cos⁡θ−1=−(1−cos⁡θ)\cos\theta - 1 = -(1 - \cos\theta), the two ratios are negatives of each other, so one tends to 00 exactly when the other does.

VQuiz Questions

Problem 1 · Bowstring Over Bow

Given: in the doubled picture the bowstring measures 2sin⁡θ2\sin\theta and the bow measures 2θ2\theta — simplify the ratio bowstring÷bow\text{bowstring} \div \text{bow}.

✅ Correct! Doubling the picture changes both lengths by the same factor 22, so the ratio is untouched — it is exactly the quantity in limit B.
❌ Only one 22 cancelled. The 22 in the numerator cancels the 22 in the denominator; you cannot drop one and keep the other.
❌ Only one 22 cancelled. Both lengths were doubled, so both 22s go — the ratio is unchanged, not halved.
❌ Two different doublings. The bowstring is 2sin⁡θ2\sin\theta — twice a sine — not sin⁡2θ\sin 2\theta, the sine of twice the angle. Those are different numbers (2sin⁡0.6=1.1292\sin 0.6 = 1.129, sin⁡1.2=0.932\sin 1.2 = 0.932).
❌ Not quite. Write the ratio as 2sin⁡θ2θ\dfrac{2\sin\theta}{2\theta} and cancel the common factor 22.
Show solution

The bowstring is two copies of sin⁡θ\sin\theta, one above the axis and one below; the bow is two arcs of length θ\theta, one above and one below:

bowstringbow=2sin⁡θ2θ=sin⁡θθ\frac{\text{bowstring}}{\text{bow}} = \frac{2\sin\theta}{2\theta} = \frac{\sin\theta}{\theta}

Doubling turns half-lengths into whole ones and leaves the ratio alone. So limit B is a statement about this picture: as θ→0\theta \to 0 the bowstring divided by the bow tends to 11.

Problem 2 · Radians Are Not Optional

Given: the same picture but with θ\theta measured in degrees, so the arc cut off by the angle has length πθ180\dfrac{\pi\theta}{180} and not θ\theta — evaluate lim⁡θ→0sin⁡θθ\lim\limits_{\theta \to 0} \dfrac{\sin\theta}{\theta} under that convention.

✅ Correct! The chord still matches the arc, but the arc is no longer the number θ\theta — it is πθ/180\pi\theta/180 of it, and that factor survives the limit.
❌ The picture has changed. The proof works because θ\theta is the arc length. In degrees the arc is πθ/180\pi\theta/180, so the denominator is off by that factor.
❌ Reciprocal. sin⁡θθ=sin⁡θarc⋅arcθ\dfrac{\sin\theta}{\theta} = \dfrac{\sin\theta}{\text{arc}} \cdot \dfrac{\text{arc}}{\theta}, and arcθ=π180\dfrac{\text{arc}}{\theta} = \dfrac{\pi}{180}, not 180π\dfrac{180}{\pi}.
❌ Not zero. Numerator and denominator both shrink to 00; test θ=1°\theta = 1\degree: sin⁡1°=0.0174524\sin 1\degree = 0.0174524, and 0.0174524/1=0.01745240.0174524 / 1 = 0.0174524, not something vanishing.
❌ Not quite. Split the ratio into (chord ÷\div arc) times (arc ÷θ\div \theta).
Show solution

The bow-and-bowstring argument compares the chord with the arc, so keep the arc explicit:

sin⁡θθ=sin⁡θπθ/180⏟chord ÷ arc⋅πθ/180θ⏟=  π/180\frac{\sin\theta}{\theta} = \underbrace{\frac{\sin\theta}{\pi\theta/180}}_{\text{chord} \,\div\, \text{arc}} \cdot \underbrace{\frac{\pi\theta/180}{\theta}}_{= \;\pi/180}

Short pieces of curves are nearly straight, so the first factor still tends to 11. The second factor is the constant π/180\pi/180. Hence

lim⁡θ→0sin⁡θθ=1⋅π180=π180≈0.01745(θ in degrees)\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1 \cdot \frac{\pi}{180} = \frac{\pi}{180} \approx 0.01745 \qquad (\theta \text{ in degrees})

Check: sin⁡(0.1°)=0.0017453\sin(0.1\degree) = 0.0017453, and dividing by 0.10.1 gives 0.0174530.017453. This is why every calculus formula for the trig functions is stated in radians — in radians, and only in radians, the angle and the arc are the same number and the limit is 11.

Problem 3 · A Bow of Length 3θ3\theta

Given: the angle 3θ3\theta on the unit circle, whose half-bow is the arc 3θ3\theta and whose half-bowstring is sin⁡3θ\sin 3\theta — evaluate lim⁡θ→0sin⁡3θθ\lim\limits_{\theta \to 0} \dfrac{\sin 3\theta}{\theta} in two moves.

First: what is lim⁡θ→0sin⁡3θ/(3θ)\lim_{\theta \to 0} \sin 3\theta / (3\theta)?

Therefore: what is lim⁡θ→0sin⁡3θ/θ\lim_{\theta \to 0} \sin 3\theta / \theta?

✅ Correct! The bowstring and the bow of the angle 3θ3\theta become equal, so sin⁡3θ\sin 3\theta and 3θ3\theta become equal — and 3θ3\theta is three times θ\theta.
❌ Check the first move. Limit B says sin⁡(angle)angle→1\dfrac{\sin(\text{angle})}{\text{angle}} \to 1 for whatever angle shrinks to 00; here the angle is 3θ3\theta, and it appears in both places.
❌ Check the second move. sin⁡3θθ=3⋅sin⁡3θ3θ\dfrac{\sin 3\theta}{\theta} = 3 \cdot \dfrac{\sin 3\theta}{3\theta} — the missing factor multiplies the limit, it does not divide it.
❌ Not quite. Make the denominator match the angle inside the sine, then repair the difference with a constant.
Show solution

Move 1. Put the picture's own angle in the denominator. As θ→0\theta \to 0 the angle 3θ→03\theta \to 0 too, and for that angle the bowstring-over-bow ratio is

lim⁡θ→0sin⁡3θ3θ=1\lim_{\theta \to 0} \frac{\sin 3\theta}{3\theta} = 1

Move 2. The wanted ratio differs from that one by a constant factor:

sin⁡3θθ=3⋅sin⁡3θ3θ  ⟶  3⋅1=3\frac{\sin 3\theta}{\theta} = 3 \cdot \frac{\sin 3\theta}{3\theta} \;\longrightarrow\; 3 \cdot 1 = 3

Verify numerically at θ=0.01\theta = 0.01: sin⁡(0.03)=0.0299955\sin(0.03) = 0.0299955, and 0.0299955/0.01=2.999550.0299955 / 0.01 = 2.99955, already within 0.02%0.02\% of 33.

Problem 4 · What the Promise Was For

Given: the difference quotient for sin⁡x\sin x, expanded with the addition formula, sin⁡(x+Δx)−sin⁡xΔx=sin⁡x⋅cos⁡Δx−1Δx+cos⁡x⋅sin⁡ΔxΔx\frac{\sin(x + \Delta x) - \sin x}{\Delta x} = \sin x \cdot \frac{\cos \Delta x - 1}{\Delta x} + \cos x \cdot \frac{\sin \Delta x}{\Delta x} evaluate its limit as Δx→0\Delta x \to 0, using the two limits just proved.

✅ Correct! Limit A kills the sin⁡x\sin x term and limit B leaves the cos⁡x\cos x term untouched: sin⁡x⋅0+cos⁡x⋅1=cos⁡x\sin x \cdot 0 + \cos x \cdot 1 = \cos x. That is the promise kept.
❌ The two limits are the other way round. The factor beside sin⁡x\sin x is limit A, which tends to 00; the factor beside cos⁡x\cos x is limit B, which tends to 11.
❌ That is the derivative of cosine. The same two limits give ddxcos⁡x=−sin⁡x\dfrac{d}{dx}\cos x = -\sin x from a different expansion — but here the leading term is cos⁡x⋅sin⁡ΔxΔx\cos x \cdot \dfrac{\sin \Delta x}{\Delta x}.
❌ Both factors treated as 11. Limit B is 11, but limit A is 00 — the gap closes to nothing beside the arc, so its whole term disappears.
❌ Not quite. Substitute the two proved values, cos⁡Δx−1Δx→0\dfrac{\cos \Delta x - 1}{\Delta x} \to 0 and sin⁡ΔxΔx→1\dfrac{\sin \Delta x}{\Delta x} \to 1.
Show solution

Both bracketed factors are exactly the two limits the bow and the bowstring have just settled, with Δx\Delta x playing the part of θ\theta:

lim⁡Δx→0cos⁡Δx−1Δx=0(limit A: gap÷arc)\lim_{\Delta x \to 0} \frac{\cos \Delta x - 1}{\Delta x} = 0 \qquad \text{(limit A: gap} \div \text{arc)} lim⁡Δx→0sin⁡ΔxΔx=1(limit B: bowstring÷bow)\lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} = 1 \qquad \text{(limit B: bowstring} \div \text{bow)}

The coefficients sin⁡x\sin x and cos⁡x\cos x do not involve Δx\Delta x, so they ride along:

ddxsin⁡x=sin⁡x⋅0+cos⁡x⋅1=cos⁡x\frac{d}{dx}\sin x = \sin x \cdot 0 + \cos x \cdot 1 = \cos x

Every trig derivative rests on these two numbers, 00 and 11 — and both were read off one picture of a bow and its bowstring. Had the angle been in degrees, limit B would be π/180\pi/180 instead of 11, and the clean formula ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x would fail.

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