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

Theorem

 * $\displaystyle \lim_{x \to 0} \frac {\sin x} x = 1$

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 $\left({0 \,.\,.\, \dfrac \pi 2}\right)$.


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

From Area of a 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:
 * $\operatorname {area}\triangle OAB = \dfrac 1 2 \sin \theta$

From Area of a 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 $\theta \in \left({0 . . \dfrac \pi 2}\right)$.

Next, from Area of a 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:
 * $\operatorname {area}\triangle OAC = \dfrac 1 2 \tan \theta$

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

We now have the following triple inequality:


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

for all $\theta \in \left({0 . . \dfrac \pi 2}\right)$.

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 $(A)$, 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 $\left({-\dfrac \pi 2 \,.\,.\, 0}\right) \cup \left({0 \,.\,.\, \dfrac \pi 2}\right)$.

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

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 $\theta \in \left({-\dfrac \pi 2 \,.\,.\, 0}\right) \cup \left({0 \,.\,.\, \dfrac \pi 2}\right)$.

Inverting all terms and reversing the inequalities:


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

for all $\theta \in \left({-\dfrac \pi 2 \,.\,.\, 0}\right) \cup \left({0 \,.\,.\, \dfrac \pi 2}\right)$.

Taking to the limit:


 * $\displaystyle \lim_{\theta \to 0} \ 1 = 1$


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

So by the Squeeze Theorem:


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