Mathematics → Calculus → sin(x)/x

The Fundamental Trigonometric Limit

\lim_{x \to 0} \frac{\sin x}{x} = 1

On the Calculus page we listed this fact alongside $\lim_{x\to 2}(3x+1)=7$ as if they were equally obvious. They are not. This page explains why.

The Puzzle

Try to evaluate $\lim_{x\to 0}\sin(x)/x$ by direct substitution. You get $0/0$ — a form that tells you nothing. The limit might be 0, or 1, or 7, or it might not exist at all. Direct substitution has failed.

Now try numerically. Set your calculator to radians and compute:

x = 1.0

0.841471

x = 0.5

0.958851

x = 0.1

0.998334

x = 0.01

0.999983

x = 0.001

0.999999

x = → 0

→ 1.000000

The ratio marches steadily toward 1. But numerical evidence isn't proof — it shows us where to aim, not why we get there.

Why It's Hard

The natural instinct is to reach for L'Hôpital's Rule — differentiate the numerator and denominator separately when facing $0/0$ :

\lim_{x\to 0}\frac{\sin x}{x} \;\overset{?}{=}\; \lim_{x\to 0}\frac{\cos x}{1} = 1

It gives the right answer. But this argument is circular. L'Hôpital requires us to know that $\frac{d}{dx}\sin x = \cos x$ , and the standard proof of that derivative begins exactly here:

\frac{d}{dx}\sin x = \lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h} = \lim_{h\to 0}\frac{\sin x \cos h + \cos x \sin h - \sin x}{h}

Simplify and you need $\lim_{h\to 0}\frac{\sin h}{h}$ to proceed. We cannot use L'Hôpital here without already knowing what we're trying to prove. We need an independent argument — and that argument turns out to be geometric.

Geometric Setup

Consider a unit circle centred at the origin. Fix an angle $x \in (0,\, \pi/2)$ . Label three points:

$O = (0,\,0)$ — the origin

$A = (1,\,0)$ — where the circle meets the positive $x$ -axis

$P = (\cos x,\, \sin x)$ — the point on the circle at angle $x$

$Q = (1,\, \tan x)$ — where the ray $OP$ meets the tangent $x=1$

These four points define three regions whose areas we can compute exactly. Drag the slider to see how they relate:

Interactive — drag to change x

x = 0.700 rad (40.1°)

Triangle OAP

0.3221

Sector OAP

0.3500

Triangle OAQ

0.4211

ratio sin(x) / x

0.920311

→ 1.000000 as x → 0

Triangle OAP

\text{Area} = \tfrac{1}{2}\cdot 1 \cdot \sin x = \tfrac{\sin x}{2}

base = OA = 1, height = sin x

Sector OAP

\text{Area} = \tfrac{1}{2}r^2 x = \tfrac{x}{2}

unit circle sector formula

Triangle OAQ

\text{Area} = \tfrac{1}{2}\cdot 1 \cdot \tan x = \tfrac{\tan x}{2}

base = OA = 1, height = tan x

The containment is clear from the diagram: the green triangle fits inside the blue sector, which fits inside the amber triangle. Therefore:

\frac{\sin x}{2} \;\le\; \frac{x}{2} \;\le\; \frac{\tan x}{2}

The Squeeze

We have the area inequality. Now we apply the tool that turns it into a limit.

Theorem — Squeeze Theorem (Sandwich Theorem)

g(x) \le f(x) \le h(x)

near

a

(but not necessarily at

a

), and

\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L

, then:

\lim_{x\to a} f(x) = L

The function $f$ is squeezed between $g$ and $h$ . If both bounds converge to the same value, $f$ has no room to go anywhere else. We now apply this to our area inequality.

Divide the inequality $\sin x / 2 \le x/2 \le \tan x / 2$ through by $\sin x / 2$ (which is positive for $x \in (0, \pi/2)$ ):

1 \;\le\; \frac{x}{\sin x} \;\le\; \frac{1}{\cos x}

Take reciprocals — this reverses the inequalities:

\cos x \;\le\; \frac{\sin x}{x} \;\le\; 1

Now let $x \to 0^+$ . We know $\cos x \to 1$ (cosine is continuous and $\cos 0 = 1$ ). The upper bound is already the constant 1. So our function $\sin x / x$ is squeezed between something approaching 1 and the constant 1. The squeeze theorem delivers the result.

The argument for $x \to 0^-$ follows by symmetry: $\sin(-x)/(-x) = \sin x / x$ , so the left-sided limit equals the right-sided limit. The two-sided limit is therefore 1.

Formal Proof

Theorem — Fundamental Trigonometric Limit

\lim_{x \to 0} \frac{\sin x}{x} = 1

Proof.Without loss of generality assume

x \in (0, \pi/2)

. On a unit circle, the areas of triangle

OAP

, sector

OAP

, and triangle

OAQ

satisfy:

\frac{\sin x}{2} \;\le\; \frac{x}{2} \;\le\; \frac{\tan x}{2}

Multiply through by

2/\sin x > 0

1 \;\le\; \frac{x}{\sin x} \;\le\; \frac{1}{\cos x}

Take reciprocals (reversing inequalities, valid since all terms are positive):

\cos x \;\le\; \frac{\sin x}{x} \;\le\; 1

Since

\lim_{x\to 0^+}\cos x = 1

and

\lim_{x\to 0^+}1 = 1

, the Squeeze Theorem gives

\lim_{x\to 0^+}\sin x / x = 1

. By evenness of

\sin x / x

, the full two-sided limit is 1.□

The Payoff — Derivative of sin

We went through all of this for a reason. With $\lim_{x\to 0}\sin x/x = 1$ in hand, we can now prove the derivative of sine from first principles — no circular reasoning. We need one more limit first, which follows directly:

\lim_{x\to 0}\frac{1 - \cos x}{x} = 0

Proof: multiply numerator and denominator by $(1 + \cos x)$ to get $\frac{\sin^2 x}{x(1+\cos x)} = \frac{\sin x}{x} \cdot \frac{\sin x}{1+\cos x} \to 1 \cdot \frac{0}{2} = 0$ .

Now expand the derivative definition using the angle addition formula:

\frac{d}{dx}\sin x = \lim_{h\to 0}\frac{\sin(x+h) - \sin x}{h}

= \lim_{h\to 0}\frac{\sin x \cos h + \cos x \sin h - \sin x}{h}

= \sin x \underbrace{\lim_{h\to 0}\frac{\cos h - 1}{h}}_{=\;0} \;+\; \cos x \underbrace{\lim_{h\to 0}\frac{\sin h}{h}}_{=\;1}

= \cos x

The two limits we just proved slot in exactly where needed. The derivative of sine is cosine — derived cleanly, without circularity, from geometry.

Why this matters for machine learning

Gradient descent — the engine behind every neural network — requires differentiating loss functions with respect to weights. Activation functions like sigmoid and softmax involve exponentials and logarithms, but the theoretical foundations of differentiation all trace back to limit arguments like this one. Understanding why the derivative is defined as a limit, and how to evaluate limits that resist direct substitution, is the bedrock on which all of that machinery rests.

References

The geometric squeeze argument is old. Euler used area inequalities on the unit circle in Introductio in Analysin Infinitorum (1748) to establish trigonometric properties. The squeeze theorem in its modern form — with an explicit statement and proof — was formalized by Augustin-Louis Cauchy in his landmark Cours d'Analyse (1821), the same work that gave us the ε-δ definition of a limit.

The proof in exactly the form presented here — three nested regions on a unit circle, area inequalities, squeeze to the limit — appears in these standard references:

Michael Spivak (1967)

Calculus

4th ed., Chapter 15 — Spivak explicitly names the L'Hôpital circularity trap, which most textbooks silently commit. The most rigorous undergraduate treatment.

Tom M. Apostol (1967)

Calculus, Vol. 1

2nd ed., §2.3 — Apostol's treatment is careful and terse. The geometric argument is presented with full rigor and the squeeze theorem is proved before it is used.

James Stewart (2015)

Calculus: Early Transcendentals

8th ed., §3.3 — The most widely used calculus textbook in the world. The proof here is essentially Stewart's, made more explicit about the circularity.

Leonhard Euler (1748)

Introductio in Analysin Infinitorum

Vol. 1 (1748) — The original source of the unit-circle area approach to trigonometric limits. Euler did not yet have the squeeze theorem by name, but the geometric intuition is all here.

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