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Limθ−>0 (sin θ)/θ = 1

Consider this diagram:

A unit circle with center O has rays OA and OB forming a central angle, θ (assume it's less than π/2) and triangle OAB.  A line segment BC is drawn perpendicular to OA.  The length of line segment BC is sin θ.  The length of arc AB is θ.

The area of triangle AOB is (1/2) sin θ.
The area of sector AOB is (1/2) θ. 

Now extend the diagram by drawing a line tangent to the circle at A and extending ray OB so that the ray and the tangent intersect at D:

The area of triangle AOD is (1/2) tan θ.

Now area of triangle AOB < area of sector AOB < area of triangle AOD, which means

sin θ ≤ θ ≤ tan θ,

Now divide all sides by sin θ, and then take the reciprocal of all sides (switching the inequalities):

sin θ ≤ θ ≤ tan θ 
1 ≤ θ/sin θ ≤ 1/cos θ 
1 ≥ sin θ/θ ≥ cos θ 
cos θ ≤ sin θ/θ ≤ 1 

The three-way inequality, above, is valid for all θ between 0 and π/2.  Since cos -θ = cos θ, and sin -θ = -sin θ, it follows that this inequality is just as true for all θ between -π/2 and 0, so it is true for all non-zero θ between -π/2 and π/2.

Now, since cos 0=1, it follows by the Squeeze Theorem that limθ−>0 sin θ/θ = 1

Proof using δ-ε form of the limit

Without the squeeze theorem, the proof goes like this:

Now area of triangle AOB < area of sector AOB < area of triangle AOD, which means

sin θ < θ < tan θ,

so by subtracting sin θ from all sides,

0 < θ - sin θ < tan θ - sin θ 

Now I want the left side of this inequality to stay smaller than the right side, so I will divide the left side by θ and the right side by the smaller number, sin θ.

0 < 1 - (sin θ)/θ < 1/cos θ - 1,

So for any positive θ < π/2, the distance from the supposed limit, 1, to (sin θ)/θ  is less than 1/cos θ - 1.  A similar argument shows it is true for negative  θ > -π/2.

For any given ε, let ε = 1/cos θ - 1, and solve this for θ, giving θ = ± cos-1(1/(ε+1))

Now then take δ to be |cos-1(1/(ε+1))|, which is always smaller than π/2 because 1/(ε+1)f is positive.

Now assume 0 < |θ| < δ.

Since 1 - (sin θ)/θ is strictly increasing on the interval (0,π/2) (proof required . . . . . . ),

1 - (sin θ)/θ < 1 - (sin δ)/δ < 1/cos δ - 1 = ε, so

If 0 < |θ| < δ , it follows that |1 - (sin θ)/θ| < ε.


limθ−>0 (sin θ)/θ = 1

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Derivative of Sin

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