Single-Variable-Calculus · Unit 3 · Video 4 · Interactive Practice

Comparing Two Quantities That Both Go to Zero, and Why Radians Are Not Optional

IKey Formulas

Formula	Name	Condition
$\displaystyle\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1$	Limit B — bowstring over bow	$\theta$ in radians
$\displaystyle\lim_{\theta \to 0}\frac{1 - \cos\theta}{\theta} = 0$	Limit A — gap over arc	any angle unit
$\text{arc length} = \theta$ on the unit circle	Definition of the radian	the angle and its arc are one number
$\dfrac{d}{dx}\sin x = \cos x, \qquad \dfrac{d}{dx}\cos x = -\sin x$	The two trig derivatives	$x$ in radians only

Key Insight: Both limits are of the form $0/0$ , and that form fixes no value — what decides the answer is which quantity shrinks faster. Count $\theta$ in degrees and limit B becomes $\tfrac{\pi}{180} \approx 0.0175$ instead of $1$ , which turns the derivative of $\sin x$ into $\tfrac{\pi}{180}\cos x$ .

IITwo Quantities Going to Zero Together

Both numerators vanish along with $\theta$ ; only one of them keeps pace with it.

θ (rad) 1.0000

💡 Algebraically the gap is $1 - \cos\theta = \tfrac{1}{2}\theta^2 - \tfrac{1}{24}\theta^4 + \cdots$ — a whole power of $\theta$ smaller than the arc, which is why dividing by $\theta$ still leaves something heading to zero.

IIIThe Unit Decides the Limit

The limit is the slope of the chord from the origin, and the angle unit sets that slope.

Unit θ 1.2000

💡 Limit A survives any unit: $\dfrac{1 - \cos k\theta}{\theta} \to 0$ for every constant $k$ , since a rescaled quantity that goes to zero still goes to zero. Only limit B forces the choice of radians.

IVWhere Limit B Enters the Derivative

Limit B multiplies $\cos x$ in the derivative of $\sin x$ , so its value is the whole formula.

Step 1

Step 1 — Split the difference quotient

\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}

Step 2 — The first bracket is limit A

\frac{\cos \Delta x - 1}{\Delta x} = -\,\frac{1 - \cos \Delta x}{\Delta x} \longrightarrow 0 \quad \text{in any unit}

Step 3 — What is left is limit B

\frac{d}{dx}\sin x = \cos x \cdot \lim_{\Delta x \to 0}\frac{\sin \Delta x}{\Delta x} = \cos x \cdot B

Step 4 — In radians $B = 1$

\frac{d}{dx}\sin x = \cos x \cdot 1 = \cos x \ \checkmark

Step 5 — In degrees $B = \pi/180$

\frac{d}{dx}\sin x = \frac{\pi}{180}\cos x \approx 0.0175\,\cos x \ \neq\ \cos x

💡 The companion formula $\dfrac{d}{dx}\cos x = -\sin x$ carries exactly the same condition, and so does every rule later built on the two of them.

VQuiz Questions

Problem 1 · Which Numerator Keeps Pace

Given: $\theta = 0.02$ radians, with $\sin\theta = 0.0199987$ and $\cos\theta = 0.9998000$ — find the two ratios $\dfrac{\sin\theta}{\theta}$ and $\dfrac{1 - \cos\theta}{\theta}$ .

✅ Correct! At

\theta = 0.02

the bowstring is already

99.99\%

of the bow, while the gap is only a hundredth of the arc.

❌ The two are swapped.

\sin\theta = 0.0199987

is almost all of

\theta = 0.02

, whereas

1 - \cos\theta = 0.0002

is a two-hundredth of it.

❌ Not both. Dividing

0.0199987

0.02

gives

0.9999

, nowhere near

0.0100

❌ Not quite. Divide each numerator by

0.02

: one of them is

0.0199987

, the other is

1 - 0.9998000 = 0.0002

Show solution

Bowstring against bow:

\frac{\sin\theta}{\theta} = \frac{0.0199987}{0.02} = 0.99993

Gap against arc:

\frac{1 - \cos\theta}{\theta} = \frac{1 - 0.9998000}{0.02} = \frac{0.0002000}{0.02} = 0.01000

Both numerators are tiny, but only $\sin\theta$ is tiny in the same proportion as $\theta$ . The gap has already fallen to a hundredth of the arc, and it keeps falling: at $\theta = 0.002$ it is a thousandth.

Problem 2 · What $0/0$ Does and Does Not Tell You

Given: $\dfrac{\sin\theta}{\theta}$ and $\dfrac{1 - \cos\theta}{\theta}$ are both of the form $0/0$ as $\theta \to 0$ , yet their limits are $1$ and $0$ — choose the statement that accounts for the difference.

✅ Correct! One power of

\theta

in the numerator matches the denominator, so the ratio settles at

1

; two powers outrun it, so the ratio collapses to

0

❌ Nothing can be substituted. At

\theta = 0

both quotients read

0/0

, which carries no information at all — the quotient is informative only just next to zero.

❌ Check the sign.

\cos\theta < 1

for every small

\theta \neq 0

, so

1 - \cos\theta

is positive throughout.

❌ Not quite. Both numerators do reach

0

\theta = 0

; the limits differ because of the rate of shrinking, not the form.

Show solution

A limit of this kind compares two quantities that both go to zero, and asks how the numerator compares in size with the denominator while both are still small.

\sin\theta = \theta - \tfrac{1}{6}\theta^3 + \cdots \quad\Longrightarrow\quad \frac{\sin\theta}{\theta} = 1 - \tfrac{1}{6}\theta^2 + \cdots \to 1

1 - \cos\theta = \tfrac{1}{2}\theta^2 - \tfrac{1}{24}\theta^4 + \cdots \quad\Longrightarrow\quad \frac{1 - \cos\theta}{\theta} = \tfrac{1}{2}\theta - \cdots \to 0

Geometrically these are the two pictures: the bowstring stays almost the full length of the bow all the way down, while the gap between bowstring and curve closes far faster than the arc shortens.

Problem 3 · The Same Limit in a Different Unit

Given: gradians cut the full turn into $400$ equal parts — find $\displaystyle\lim_{\theta \to 0}\frac{\sin\theta}{\theta}$ with $\theta$ counted in gradians.

✅ Correct! An angle of

\theta

gradians cuts an arc of length

\tfrac{2\pi}{400}\theta = \tfrac{\pi}{200}\theta

, so the ratio is

\tfrac{\pi}{200}

times bowstring over bow.

❌ That is the radian answer. It holds only because a radian angle and the arc it cuts off are the same number, which is false in gradians.

❌ That is the degree factor.

\tfrac{2\pi}{360} = \tfrac{\pi}{180}

divides the turn into

360

parts; gradians divide it into

400

❌ That is the reciprocal. The arc is

\tfrac{\pi}{200}\theta

, so the ratio is multiplied by

\tfrac{\pi}{200}

, not divided by it.

❌ Not quite. Start from the circumference: the unit circle has circumference

2\pi

, and

\theta

gradians is

\theta/400

of it.

Show solution

Step 1 — Find the arc that $\theta$ gradians cuts off. The unit circle has circumference $2\pi$ , split into $400$ equal parts:

\text{arc} = \frac{2\pi}{400}\,\theta = \frac{\pi}{200}\,\theta

Step 2 — The geometry is unchanged. Bowstring over bow still tends to $1$ , so with $\alpha = \tfrac{\pi}{200}\theta$ the radian arc,

\frac{\sin\theta}{\theta} = \frac{\sin\alpha}{\alpha}\cdot\frac{\alpha}{\theta} = \frac{\sin\alpha}{\alpha}\cdot\frac{\pi}{200}

Step 3 — Take the limit. As $\theta \to 0$ so does $\alpha$ , and $\dfrac{\sin\alpha}{\alpha} \to 1$ :

\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = \frac{\pi}{200} \approx 0.015708

Radians are the one unit in which this constant is $1$ , because there — and only there — the angle and its arc are the same number.

Problem 4 · The Derivative of Sine in Degrees

Given: with $x$ measured in degrees the same algebra still holds, $\frac{\sin(x + \Delta x) - \sin x}{\Delta x} = \sin x \cdot \frac{\cos \Delta x - 1}{\Delta x} + \cos x \cdot \frac{\sin \Delta x}{\Delta x}$ — find the limit of the first bracket and the derivative formula that results.

What does the first bracket tend to?

So with $x$ in degrees, what is $\dfrac{d}{dx}\sin x$ ?

✅ Correct! Only the second bracket carries the unit, and in degrees it settles at

\tfrac{\pi}{180}

instead of

1

— so the familiar formula is simply wrong there.

❌ Check the first bracket. It is

-\dfrac{1 - \cos \Delta x}{\Delta x}

, which is limit A — and limit A goes to

0

in every unit, because changing the unit only rescales a quantity that is already heading to zero.

❌ Check the second bracket. It is limit B, worth

\tfrac{\pi}{180}

in degrees, and it multiplies

\cos x

Show solution

Step 1 — The first bracket. In degrees, $\Delta x$ degrees is $\tfrac{\pi}{180}\Delta x$ radians, so

\frac{\cos \Delta x - 1}{\Delta x} = -\,\frac{1 - \cos\!\big(\tfrac{\pi}{180}\Delta x\big)}{\Delta x} \approx -\,\frac{\tfrac{1}{2}\big(\tfrac{\pi}{180}\Delta x\big)^2}{\Delta x} \to 0

The $\sin x$ term therefore drops out, exactly as it does in radians.

Step 2 — The second bracket. This one is limit B, and it is the bracket the unit reaches:

\lim_{\Delta x \to 0}\frac{\sin \Delta x}{\Delta x} = \frac{\pi}{180} \approx 0.017453

Step 3 — Put them together.

\frac{d}{dx}\sin x = \sin x \cdot 0 + \cos x \cdot \frac{\pi}{180} = \frac{\pi}{180}\cos x

In radians the same two steps give $\cos x \cdot 1 = \cos x$ . The clean formula exists only because limit B is exactly $1$ , and that happens only when the angle is measured in radians.

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