 Limits of Trigonometric Functions How to solve the limits of trigonometric functions problems: the limit of (sin x)/x as x goes to 0, proof, examples, and their solutions.

Formula The limit of (sin x)/x as x → 0 is 1.

This formula is mostly used
when solving the limits of trigonometric functions problems.

Proof To prove this limit formula,
you should show that

the limit of (sin x)/x as x → 0+
and
the limit of (sin x)/x as x → 0-

are equal.

One-sided limits

Case 1: as x → 0+

On a unit circle,
draw a terminal side whose central angle is x (rad).

Draw △AOB (purple), sector AOB (blue), and △AOT (green).

The y value of point T is tan x.
(∴ Tangent: TOA
So, in △AOT, tan x = [y value]/1.)

The area of △AOB is (sin x)/2.

Area of a triangle (using sine)

The area of sector AOB is x/2.

Area of a circular sector (in radian)

And the area of △AOT is (tan x)/2.

See the figure above.

It's obvious that
(△AOB) < (Sector AOB) < (△AOT).

So x/2 < (sin x)/2 < (tan x)/2.

Divide each side by 2.

Change tan x = (sin x)/(cos x).

Quotient identities

Divide each side by sin x.

Switch each side's numerator and the denominator.

Each side is (+).
So the orders of the inequality signs do change.

Set the limit on each side.

As x → 0+,
the left side, 1, goes to 1,
and the right side, cos x, also goes to 1.

So, by the squeeze theorem,
the limit of (sin x)/x as x → 0+ is 1. Case 2: as x → 0-

Set x = -t.

Then, as x → 0-,
t → 0+.

Put -t into the x-s.

x → 0- has to change into t → 0+.

sin (-t) = -sin t

Trigonometric functions of (-θ)

By case 1,
(limit) = 1.

Example 1 To make 4x on the denominator,
multiply 4 to the denominator's x.

And, to undo the 4,
multiply 4 to the numerator.

Then, as x → 0,
the limit of (sin 4x)/(4x) is 1.

So (limit) = 1⋅4 = 4.

Example 2 As x → 0,
the limit of (sin x)/x is 1,
and the limit of 1/cos x is 1/1.

Graphing cosine functions

Example 3 Make sin x (gray) on the denominator.

And to undo the sin x (gray),
write sin x on the numerator
with the denominator x.

Then, as x → 0,
the limit of [sin (sin x)]/(sin x) is 1,
and the limit of (sin x)/x is 1.

So (limit) = 1⋅1 = 1.

Example 4 To solve limit problems with (1 - cos x),
multiply its conjugate, (1 + cos x),
to both of the numerator and the denominator.

Then (1 - cos x - 1)(1 + cos x) = 1 - cos2 x.

This method also can be used
when solving limit problems with (1 + cos x).

1 - cos2 x = sin2 x

Pythagorean identities

As x → 0,
the limit of (sin2 x)/(x2) is 12,
and the limit of 1/(1 + cos x) is 1/(1 + 1).

So (limit) = 12⋅(1/(1 + 1)) = 1/2.