The Limit of sin x over x

Substituting gives 0 over 0

At x = 0, sin x is 0 and x is 0, so $\frac{\sin x}{x}$(sin x)/x reads $\frac{0}{0}$0/0. That is an indeterminate form: it decides nothing, and the limit as $x\to 0$x → 0 has to be found another way.

Here there is no factor to cancel. sin x is not x times anything simpler, so the rewriting that worked for $\frac{x^2-4}{x-2}$(x² − 4)/(x − 2) has nothing to work on.

The values near 0

With x in radians, $\frac{\sin x}{x}$(sin x)/x is 0.8415 at x = 1, 0.9589 at x = 0.5, 0.99833 at x = 0.1, and 0.999983 at x = 0.01. The values climb toward 1.

The other side gives the same values, because sin(−x) = −sin x, so $\frac{\sin \left(-x\right)}{-x}=\frac{\sin x}{x}$(sin(−x))/(−x) = (sin x)/x. The graph is symmetric about the y-axis, and from both sides it closes in on height 1. At x = 0 itself there is no value, so the graph has a hole at (0, 1).

The graph of $y=\frac{\sin x}{x}$y = (sin x)/x for x from −6 to 6, in radians. It is 0 where sin x is, at $x=\pm \pi$x = ±π, and rises to a hollow dot at (0, 1): the height the curve approaches, at the one point where it has no value.

Three areas on the unit circle

Take an angle x with $0<x<\frac{\pi }{2}$0 < x < π/2, at the center O of a circle of radius 1. Let A = (1, 0), let P = (cos x, sin x) be the point at angle x on the circle, and let T be where the line OP meets the tangent to the circle at A, the vertical line through A. Then T = (1, tan x), because OAT is a right triangle with OA = 1 and angle x at O.

Three regions sit inside one another. The triangle OAP has base OA = 1 and height sin x, so its area is $\left(\frac{1}{2}\right)\sin x$(1/2) sin x. The sector OAP has area $\left(\frac{1}{2}\right)\times 1^2\times x=\frac{x}{2}$(1/2) × 1² × x = x/2, a formula that holds only with x in radians. The triangle OAT has base 1 and height tan x, so its area is $\left(\frac{1}{2}\right)\tan x$(1/2) tan x.

The triangle OAP lies inside the sector, and the sector lies inside the triangle OAT. So $\left(\frac{1}{2}\right)\sin x<\frac{x}{2}<\left(\frac{1}{2}\right)\tan x$(1/2) sin x < x/2 < (1/2) tan x, and doubling, sin x < x < tan x.

The angle x = 0.8 radians at O. The shaded sector OAP has area $\frac{x}{2}=0.4$x/2 = 0.4. The dashed triangle OAP inside it has area $\left(\frac{1}{2}\right)\sin 0.8=0.359$(1/2) sin 0.8 = 0.359, and the dashed triangle OAT, with T on the tangent line at A, has area $\left(\frac{1}{2}\right)\tan 0.8=0.515$(1/2) tan 0.8 = 0.515.

The squeeze

Divide sin x < x < tan x by sin x, which is positive for $0<x<\frac{\pi }{2}$0 < x < π/2: $1<\frac{x}{\sin x}<\frac{1}{\cos x}$1 < x/(sin x) < 1/(cos x). All three are positive, and taking reciprocals of positive numbers reverses their order: $\cos x<\frac{\sin x}{x}<1$cos x < (sin x)/x < 1.

For $-\frac{\pi }{2}<x<0$−π/2 < x < 0 the same holds, because cos(−x) = cos x and $\frac{\sin x}{x}$(sin x)/x takes the same value at −x as at x. So near 0, on both sides, $\frac{\sin x}{x}$(sin x)/x is trapped between cos x below and 1 above.

As $x\to 0$x → 0, $\cos x\to \cos 0=1$cos x → cos 0 = 1, and the constant 1 stays at 1. Both bounds tend to 1, so by the squeeze theorem $\frac{\lim \left(\sin x\right)}{x}$lim (sin x)/x as $x\to 0$x → 0 is 1.

From x = −1.5 to 1.5: the gold curve $y=\frac{\sin x}{x}$y = (sin x)/x between the dashed curve y = cos x below and the dashed line y = 1 above. At x = 1.5 they are 0.071, 0.665 and 1; toward x = 0 all three meet at height 1.

Why radians

The area of a sector is $\frac{x}{2}$x/2 only when x is in radians, and the whole argument rests on it. In degrees the answer changes. An angle of x degrees is $\frac{\pi x}{180}$πx/180 radians, so $\frac{\sin \left(x^{\circ }\right)}{x}=\left(\frac{\pi }{180}\right)\times \frac{\sin \left(\frac{\pi x}{180}\right)}{\frac{\pi x}{180}}$sin(x°)/x = (π/180) × sin(πx/180)/(πx/180), which tends to $\frac{\pi }{180}=0.017453$π/180 = 0.017453 as $x\to 0$x → 0. At x = 1 it is already 0.017452.

A limit of 1 says that near 0, sin x and x are nearly the same number: at x = 0.1, sin x = 0.09983. In radians the sine curve leaves the origin running alongside the line y = x. The chord from (0, 0) to (h, sin h) has gradient $\frac{\sin h}{h}$(sin h)/h, which tends to 1, so the gradient of the sine curve at the origin is 1. Working out the gradient of sin x at every point rests on this same limit, and in degrees every such result would carry an extra factor of $\frac{\pi }{180}$π/180.

Limits built from this one

For $\frac{\sin 3x}{x}$(sin 3x)/x as $x\to 0$x → 0, write it as $3\times \frac{\sin 3x}{3x}$3 × (sin 3x)/(3x). As $x\to 0$x → 0, $3x\to 0$3x → 0 as well, so $\frac{\sin 3x}{3x}\to 1$(sin 3x)/(3x) → 1 and the limit is 3. At x = 0.01 the value is 2.99955.

For $\frac{\tan x}{x}$(tan x)/x, write tan x as $\frac{\sin x}{\cos x}$(sin x)/(cos x): $\frac{\tan x}{x}=\left(\frac{\sin x}{x}\right)\times \frac{1}{\cos x}\to 1\times 1=1$(tan x)/x = ((sin x)/x) × 1/(cos x) → 1 × 1 = 1. At x = 0.01 the value is 1.000033.

For $\frac{\sin 5x}{\sin 2x}$(sin 5x)/(sin 2x), write it as $\left(\frac{\sin 5x}{5x}\right)\times \left(\frac{2x}{\sin 2x}\right)\times \frac{5}{2}$((sin 5x)/(5x)) × ((2x)/(sin 2x)) × 5/2. The first two factors tend to 1, so the limit is $\frac{5}{2}$5/2. At x = 0.01 the value is 2.4991.

The usual mistakes

Answering 0 because sin 0 = 0. The denominator tends to 0 as well, and $\frac{0}{0}$0/0 decides nothing.

Canceling the x. $\frac{\sin x}{x}$(sin x)/x is not sin, and sin x is not a product with a factor x to cancel.

Working in degrees. Then $\frac{\sin x}{x}$(sin x)/x tends to $\frac{\pi }{180}$π/180, not 1, and the approximation $\sin x\approx x$sin x ≈ x fails.

Bounding the ratio by −1 and 1, or by 0 and 1. Those bounds do not share a limit, so they force nothing. The bounds that close together are cos x and 1.

Reading the limit as an equality. $\frac{\sin x}{x}$(sin x)/x is less than 1 for every $x\ne 0$x ≠ 0; it only approaches 1.

A satellite’s chord and arc

In the application below, the straight-line distance between two points on an orbit, divided by the distance along the orbit, comes down to $\frac{\sin u}{u}$(sin u)/u with u half the angle between them, so its limit is this one.