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
Asector = (Φ/(2π)) π r2 = 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
A013 = ( ) ( ) / 2 = sin(Φ) / 2
The area of the right triangle is
A023 = ( ) ( ) / 2 = tan(Φ) / 2 = sin(Φ) / (2 cos(Φ))
Looking at the diagram above, the areas can be arranged in order as
A013 < Asector < A023
and this can be rewritten as
sin(Φ) / 2 < Φ/2 < tan(Φ) / 2
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.