linguistic:voice_change, information:classification
θ as θ tends to 0

The diagram at right shows a circle with centre O and radius r = 1. Let two radii OA and OB make an arc of θ radians. Since we are considering the limit as θ tends to zero, we may assume θ is a small positive number, say 0 < θ < 1 / 2 π in the first quadrant.

In the diagram, let R be the triangle OAB, R the circular sector OAB, and R the triangle OAC.

The area of triangle OAB is:

The area of the circular sector OAB is:

The area of the triangle OAC is given by:

Since each region is contained in the next, one has:

Moreover, since sin θ > 0 in the first quadrant, we may divide through by 1 / 2 sin θ, giving:

In the last step we took the reciprocals of the three positive terms, reversing the inequities.

We conclude that for 0 < θ < 1 / 2 π, the quantity sin(θ)/ θ is always less than 1 and always greater than cos(θ). Thus, as θ gets closer to 0, sin(θ)/ θ is " squeezed " between a ceiling at height 1 and a floor at height cos θ, which rises towards 1; hence sin(θ)/ θ must tend to 1 as θ tends to 0 from the positive side:

lim θ → 0 + sin θ θ = 1. {\displaystyle \lim _{\theta \to 0^{+}}{\frac {\sin \theta }{\theta }}=1\,.}

For the case where θ is a small negative number – 1 / 2 π < θ < 0, we use the fact that sine is an odd function: