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 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  <  A_sector  <  A₀₂₃

 

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.