Proofiness - Limit of sin(x)/x

This is my proof for the limit of [sin(x)/x] as x approaches 0 equals 1. I recommend that you print out the image, and mark it as you read the proof, this should help you in understanding the proof better, and make it 

easier to follow along.

Given: A unit circle on the cartesian plane, with circle A. Hence line segments AB and AD are 1 unit in length.

Construction:
1) BC is a prependicular through point B onto the X-axis. Consequently angle BAC lies in the first quadrant.
2) A line segment joining points B and D, hereinafter referred to as line segment BD.
3) A tangent to circle A at point D, marked by line DE.
4) Extend line DE and AB and mark their point of interesection as point F

To prove: limit [sin(x)/x] as x approaches 0 equals 1.

Proof:
Assume, angle BAC is equal to x degrees.
Now, in triangle BAC:
sin(x) = BC/AB = BC/1 (As circle A is a unit circle, and AB is its radius)

sin(x) = BC - I

Now, in triangle AFD:
tan(x) = FD/AD = FD/1 (As circle A is a unit circle, and AD is its radius)

tan(x) = FD - II

Area of triangle ABD = 1/2 x (BC) x (AD) = 1/2 x (sin(x)) x (1) = 0.5sin(x) - From statement I

Area of sector ABD = 1/2 x Angle BAC x (AD) = 1/2 x (x) x (1) = 0.5(x) - Area of a sector

Area of triangle AFD = 1/2 x (FD) x (AD) = 1/2 x (tan(x)) x (1) = 0.5 tan(x) - From statement II

Now, we can observe that:

Area of triangle AFD > Area of sector ABD > Area of triangle ABD

0.5tan(x) > 0.5(x) > 0.5 sin(x)

Multiply throughout by 2

tan(x) > x > sin(x)

Now, we know that pi/2 > angle BAC = x > 0

Hence, tan(x) > 0 | x > 0 | and sin(x) > 0

Using laws of positive numbers and inequalities:

1/tan(x) < 1/x < 1/ sin(x)

Multiply throughout by sin(x)

1/cos(x) < sin(x)/x < 1

Now, limit of 1/cos(x) as x approaches 0 is 1 and limit of 1 as x approaches 0 is 1

Thus, by using the squeeze theorem for continuous functions

limit of [sin(x)/x] as x approaches 0 is 1

QED