The limits of sin x/x at zero and infinity, with a noncircular proof

Two different limits

With $x$ measured in radians,

$$\boxed{\lim_{x\to0}\frac{\sin x}{x}=1,\qquad\lim_{x\to\infty}\frac{\sin x}{x}=0.}$$

The first concerns the local behavior of sine near zero. The second concerns a bounded numerator divided by a quantity growing without bound. They require different squeezes.

The geometric proof near zero

For $0<x<\pi/2$, compare a triangle inside a unit-circle sector with the sector and a triangle outside it. In the figure, $A$ is the circle's center, $AB=1$, and $BD$ is tangent to the circle at $B$.

Their areas give

$$\frac12\sin x<\frac x2<\frac12\tan x.$$

Because $x>0$ and $\cos x>0$, these inequalities imply

$$\cos x<\frac{\sin x}{x}<1.$$

As $x\to0^+$, both bounds tend to one. The squeeze theorem gives the right-hand limit $1$.

For negative $x$, use the fact that $\sin(-x)/(-x)=\sin x/x$. The quotient is even, so the left-hand limit is the same. This proves the two-sided limit.

Why radians matter

The unit-sector area is half the angle only when the angle is measured in radians. If a variable $d$ instead measures degrees, then $x=\pi d/180$, so

$$\lim_{d\to0}\frac{\sin(\pi d/180)}d=\frac\pi{180}.$$

The denominator's units cannot be ignored.

The limit at infinity

For $x>0$,

$$0\le\left|\frac{\sin x}{x}\right|\le\frac1x.$$

The upper bound tends to zero as $x\to\infty$, hence

$$\boxed{\lim_{x\to\infty}\frac{\sin x}{x}=0.}$$

The same absolute-value argument also proves the limit zero as $x\to-\infty$, using $1/|x|$ as the upper bound. An inequality written with $1/x$ as a nonnegative upper bound applies only when $x>0$.

Is the sector-area proof circular?

It depends on which geometric facts have already been established independently. One possible circular chain is:

1. Use the sine limit to prove the derivatives of sine and cosine.
2. Use those derivatives in a trigonometric substitution to compute a circular sector's area.
3. Use that sector-area formula to prove the same sine limit.

If sector area and radians have been developed geometrically without those derivative formulas, the usual proof has no such circularity. The argument is not automatically invalid just because a sector appears in it.

Here is an alternative development using only algebraic differentiation, integration by parts, and an independently established arc-length formula.

Establish sector area without trigonometric substitution

Choose $0<s<1$ on the right-hand unit semicircle $x=\sqrt{1-y^2}$. Let

$$I(s)=\int_0^s\sqrt{1-y^2}\,dy.$$

The sector bounded by the positive horizontal axis and the radius through $(\sqrt{1-s^2},s)$ has area

$$A(s)=I(s)-\frac12s\sqrt{1-s^2}.$$

Instead of substituting a sine or cosine, integrate by parts with the differentiated factor $y$:

$$I(s)=\left[y\sqrt{1-y^2}\right]_0^s+\int_0^s\frac{y^2}{\sqrt{1-y^2}}\,dy.$$

Since $y^2=(y^2-1)+1$, the remaining integral is

$$-I(s)+\int_0^s\frac1{\sqrt{1-y^2}}\,dy.$$

Thus

$$I(s)=\frac12s\sqrt{1-s^2}+\frac12\int_0^s\frac1{\sqrt{1-y^2}}\,dy.$$

Subtract the triangle area:

$$A(s)=\frac12\int_0^s\frac1{\sqrt{1-y^2}}\,dy.$$

Identify the integral as unit-circle arc length

For the same algebraic curve,

$$\frac{dx}{dy}=-\frac y{\sqrt{1-y^2}}.$$

Its arc length from height zero to height $s$ is

$$L(s)=\int_0^s\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy=\int_0^s\frac1{\sqrt{1-y^2}}\,dy.$$

Therefore $A(s)=L(s)/2$. Radian measure on a unit circle is its arc length, so if the central angle is $\theta=L(s)$, the sector has area $\theta/2$.

No derivative of a sine or cosine was used. We can now use the sector-area comparison in the squeeze proof without assuming the sine limit along the way. This construction assumes an arc-length theorem for smooth algebraic curves that has itself been established independently of trigonometric differentiation.

The integration-by-parts step is an example of a useful general tactic: when substitution would depend on a theorem not yet available, a different algebraic identity may avoid that dependency.

Related: three useful limit formulas and integration by parts.