# 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).

Radian measure

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 4*x* 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 4*x*)/(4*x*) 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 - cos^{2} *x*.

This method also can be used

when solving limit problems with (1 + cos *x*).

1 - cos^{2} *x* = sin^{2} *x*

Pythagorean identities

As *x* → 0,

the limit of (sin^{2} *x*)/(*x*^{2}) is 1^{2},

and the limit of 1/(1 + cos *x*) is 1/(1 + 1).

So (limit) = 1^{2}⋅(1/(1 + 1)) = 1/2.