The limit

lim x→0 (sin x / x) = 1

is one of the most important limits in calculus. It appears repeatedly when finding derivatives of trigonometric functions and in many other fundamental results.

A particularly elegant way to understand this limit is through the geometry of a unit circle.

1. Start with a unit circle

Consider a circle with centre O and radius 1. Let A be the point where the circle meets the positive x-axis, and let C be another point on the circle corresponding to a small positive angle x.

Because the radius is 1:

OA = OC = 1

The angle AOC is x radians.

For this small angle, three lengths can be compared:

• the vertical side of the inscribed triangle,
• the circular arc,
• the corresponding tangent segment.

These produce the crucial inequality.

---

2. The inscribed triangle gives sin x

Join O to C and draw a perpendicular from C to the x-axis.

This creates a right triangle whose hypotenuse is OC, which has length 1.

By the definition of sine:

sin x = opposite side / hypotenuse

Since the hypotenuse is 1, the opposite side has length:

sin x

Thus, the vertical side of the triangle represents sin x.

---

3. The arc has length x

The arc between A and C is part of the unit circle.

The length of an arc is:

arc length = radius × angle

when the angle is measured in radians.

Since the radius is 1:

arc length = 1 × x = x

Therefore, the arc AC has length x.

This is why radians are essential in this proof. For a unit circle, the numerical value of an angle in radians is equal to the length of the corresponding arc.

---

4. The tangent triangle gives tan x

Now draw a tangent to the circle at A.

The tangent is perpendicular to OA. Extend OC until it meets the tangent at a point B.

Triangle OAB is a right triangle.

Since OA = 1:

tan x = AB / OA

Therefore:

AB = tan x

So the tangent segment has length tan x.

---

5. Comparing the three lengths

For a small positive angle x, the geometry gives:

sin x < x < tan x

This is the central inequality of the proof.

The straight chord is shorter than the curved arc, while the tangent segment extends beyond the arc.

Therefore:

sin x < x < tan x

---

6. Rearranging the inequality

Starting with:

sin x < x < tan x

divide all three parts by sin x:

1 < x / sin x < tan x / sin x

Since:

tan x = sin x / cos x

the right-hand side becomes:

1 / cos x

Therefore:

1 < x / sin x < 1 / cos x

Taking reciprocals reverses the inequalities:

1 > sin x / x > cos x

or:

cos x < sin x / x < 1

This is exactly the form needed for the Squeeze Theorem.

---

7. Let x approach zero

Now consider what happens as x gets closer and closer to zero.

The lower boundary is:

cos x

and the upper boundary is:

1

As x approaches zero:

cos x → 1

The upper boundary is already 1.

Therefore:

cos x < sin x / x < 1

becomes increasingly squeezed between two quantities approaching 1.

By the Squeeze Theorem:

lim x→0 (sin x / x) = 1

---

The complete argument in one chain

The geometric proof can be remembered through three steps:

sin x < x < tan x

which leads to:

cos x < sin x / x < 1

and finally:

lim x→0 (sin x / x) = 1

Why this proof is important

Direct substitution into sin x / x gives:

0 / 0

which is an indeterminate form.

Instead of trying to evaluate 0/0 directly, geometry shows that the ratio is trapped between two quantities that both approach 1.

The ratio therefore has no choice but to approach 1.

The remarkable part of this proof is that a fundamental calculus limit emerges from the simple geometry of a unit circle, an inscribed triangle, an arc, and a tangent line.