Limit of Sine of X over X at Zero/Geometric Proof

Geometric Proof
Let $\theta$ be an angle in the unit circle, measured in radians.

Define the following points:

and consider all $\theta$ in the open interval $\openint 0 {\dfrac \pi 2}$.


 * Limit of Sine of X over X-Proof 3.png

From Area of Triangle in Terms of Side and Altitude, we have that $\triangle OAB$ has an area of $\dfrac 1 2 b h$ where:
 * $b = 1$
 * $h = \sin \theta$

and so:
 * $\Area \triangle OAB = \dfrac 1 2 \sin \theta$

From Area of Sector, the sector formed by arc $AB$ subtending $O$ is $\dfrac \theta 2$.

Clearly this sector cannot be smaller in area than $\triangle OAB$, and so we have the inequality:
 * $\dfrac {\sin \theta} 2 \le \dfrac \theta 2$

for all $\theta \in \openint 0 {\dfrac \pi 2}$.

Next, from Area of Triangle in Terms of Side and Altitude, we have that $\triangle OAC$ has an area of $\dfrac 1 2 b h$ where:
 * $b = 1$
 * $h = \tan \theta$

and so:
 * $\Area \triangle OAC = \dfrac 1 2 \tan \theta$

$\triangle OAC$ is clearly not smaller than the sector.

We now have the following compound inequality:


 * $(\text A) \quad \dfrac 1 2 \sin \theta \le \dfrac 1 2 \theta \le \dfrac 1 2 \tan \theta$

for all $\theta \in \openint 0 {\dfrac \pi 2}$.

Clearly, if any of the plane regions were to be reflected about the $x$-axis, the magnitudes of the signed areas would be the same.

The inequality $(\text A)$, then, will hold in quadrant $\text{IV}$ if the absolute value of all terms is taken, and so:

Now, we have that $\dfrac \theta {\sin\theta} \ge 0$ on $\openint {-\dfrac \pi 2} 0 \cup \openint 0 {\dfrac \pi 2}$.

Also, we have that $\dfrac 1 {\cos \theta} \ge 0$ on $\openint {-\dfrac \pi 2} 0 \cup \openint 0 {\dfrac \pi 2}$.

So our absolute value signs are not needed.

Hence we arrive at:


 * $1 \le \dfrac \theta {\sin \theta} \le \dfrac 1 {\cos \theta}$

for all $\theta \in \openint {-\dfrac \pi 2} 0 \cup \openint 0 {\dfrac \pi 2}$.

Inverting all terms and reversing the inequalities:


 * $1 \ge \dfrac {\sin\theta} \theta \ge \cos \theta$

for all $\theta \in \openint {-\dfrac \pi 2} 0 \cup \openint 0 {\dfrac \pi 2}$.

Taking to the limit:


 * $\ds \lim_{\theta \mathop \to 0} 1 = 1$


 * $\ds \lim_{\theta \mathop \to 0} \cos \theta = 1$

So by the Squeeze Theorem:


 * $\ds \lim_{\theta \mathop \to 0} \frac {\sin \theta} \theta = 1$