The tutor shows a reason for the famous limit _x→0 sinx/x = 1.

The area of a sector of a circle about angle Ɵ is Ɵr²/2. (A sector is a pie slice.) The area of a triangle is base(height)/2.

In the above drawing, the angle is x. The radius of the circle is 1. The area of the inner sector is x(cosx)²/2. The area of the triangle is cosx(sinx)/2. The area of the outer sector is x(1)²/2 = x/2.

Visually, the area of the inner sector is less than (or equal) that of the triangle, which in turn is less than (or equal) that of the outer sector. Therefore,

x(cosx)²/2 ≤ cosx(sinx)/2 ≤ x/2

Multiplying by 2 gives

x(cosx)² ≤ cosx(sinx) ≤ x

Now, dividing by x gives

(cosx)² ≤ cosx(sinx)/x ≤ x/x

which simplifies to

(cosx)² ≤ cosx(sinx)/x ≤ 1

Now, let’s imagine

lim_x→0(cosx)² ≤ lim_x→0cosx(sinx)/x ≤ lim_x→01

Cos(0)=1; of course, lim_x→01 = 1. Hence, we have

1 ≤ lim_x→0(sinx)/x ≤ 1

which means

lim_x→0(sinx)/x = 1 (Squeeze Theorem).

HTH:)

Source:

Larson, Roland and Robert Hostetler. Calculus, 3rd ed. Toronto: D C Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.