**Limit of Sin(****Φ****) / ****Φ**

The limit [ sin(Φ) / Φ ] as Φ->0, is at first glance unknown as both the numerator and denominator

go to zero. And division by zero is always undefined. Nevertheless the result of the division does

approach a finite value. To derive this result, consider the following drawing.

The above circle has a radius of one so that

The right triangle has a height of = r sin(Φ) = sin(Φ)

The right triangle has a height of = r tan(Φ) = sin(Φ) / cos(Φ)

The area of that fraction of the circle in the “sector” (not the triangle) is

A_{sector} = (Φ/(2π))
π r^{2 }=^{ }Φ/2

where the angle Φ is expressed in radians and the area includes that portion between the

circumference of the circle and the chord .

The area of the isosceles triangle is

A_{013} = ( ) ( ) / 2 = sin(Φ) / 2

The area of the right triangle is

A_{023} = ( ) ( ) / 2 = tan(Φ) / 2 = sin(Φ)
/ (2 cos(Φ))

Looking at the diagram above, the areas can be arranged in order as

A_{013} < A_{sector}
< A_{023}

and this can be rewritten as

sin(Φ) / 2 < Φ/2 < tan(Φ) / 2

or

1 < Φ / sin(Φ) < 1 / cos(Φ)

or finally by inverting each term and thus reversing the order

cos(Φ) > sin(Φ) / Φ > 1

And because in the “limit” as the angle Φ gets smaller and smaller, we can write

limit(Φ->0) [ cos(Φ) ] = 1

then the term in the middle is “squeezed” to have the same value as at both ends or

**limit(****Φ****->0)****
[ sin(****Φ) / ****f**** ] = 1**

which is what we finally wanted to show, namely that the ratio approaches a finite value in the limit.

And this is an important result especially for the derivatives of trigonometric functions.