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

Theorem

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

Corollary

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

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


 * [[File:Limit of Sine of X over X-Proof 3.png]]

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

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

From Area of a Sector, the sector formed by arc $BC$ subtending $A$ is $\dfrac \theta 2$.

Clearly this sector cannot be smaller in area than $\triangle ABC$, 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 $\triangle ABD$ has an area of $\dfrac 1 2 b h$ where:
 * $b = 1$
 * $h = \tan \theta$

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

$\triangle ABD$ 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 $\theta \in \left({0 . . \dfrac \pi 2}\right)$.

If any of the plane regions are reflected about the $x$-axis, the magnitudes of the signed areas would be the same.

The inequality $(A)$, then, will hold in quadrant 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:


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

Proof of Corollary

We have the inequality


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

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

Taking the limit of the leftmost term and the rightmost term:


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


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

So by the Squeeze Theorem:


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