To show that the limit of \(sin\theta \over \theta\) \(\approx 1\) as \(\theta \to 0\) can be shown by the Squeeze Theorem. The figure above is a unit circle with two similar triangles inscribed on it.
See above figure.
We start by letting the side adjacent of the angle \(\theta\) of the smaller triangle be of length \(a\), the side opposite, be of length \(b\) and the hypotenuse is on the unit circle and is therefore of length 1. This gives us the following:
Area of smaller triangle = \(ab \over 2\)
The area of the arc is the area of that part of the circle. If angle \(\theta\) is in radians, the area of the arc is \(\theta \over 2\pi\)\(\pi\)\(r^2\). Sinse the radius is 1 this can be simplified to Area = \(\theta \over 2\).
Area of arc = \(\theta \over 2\pi\)\(\pi\)\(1^2\) simplifies to \(\theta \over 2\).
The area of the larger triangle by the following: Let the side opposite of the angle \(\theta\) be of length \(c\) and since the side adjacent is on the unit circle, it is of length \(1\). This gives us an area of the larger triangle to be \(c \over 2\).
Area of larger triangle = \(c \over 2\).
It should be clear by the figure that the smaller triangle has an area less than or equal to the area of the arc, and that the area of the arc is less than or equal to the area of the larger triangle. Therefore we can set up the following inequality:
$${ab \over 2} {\le} {\theta \over 2} {\le} {c \over 2}$$
Multiplying everything by two, we get:
$${ab \le \theta \le c}$$
Since the hypotenuse of the smaller triangle is 1. We can replace \(ab\) with \(coshsin\theta\). Similarly, since the side adjacent of the larger triangle is 1, \(tan\theta\) can replace \(c\) and, in turn, can be replaced by \(sin\theta \over cos\theta\). Thus we get:
$${cos\theta sin\theta}{\le}{\theta}{\le}{sin\theta \over cos\theta}$$
Dividing everything by \(sin\theta\), we get:
$${cos\theta}{\le}{\theta \over sin\theta}{\le}{1 \over cos\theta}$$
But we're not solving for \(\theta \over sin\theta\), we're looking for \(sin\theta \over \theta\). We do this by inverting everything and reversing the inqualities to get:
$${1 \over cos\theta}{\ge}{sin\theta \over \theta}{\ge}{cos\theta}$$
As \(\theta \to 0\), \(1 \over cos\theta\) is decreasing toward a value of one. Also, \(cos\theta\) is increasing toward a value of one. So by the Squeeze theorem, \(sin\theta \over \theta\) \(\approx 1\). QED
$${sin\theta \over \theta} {\approx 1}$$