8. Geometric Pinching

Proving that

$$\lim_{x \to 0} \dfrac{\sin(x)}{x} = 1$$

where circular arc $AP$ is centered about $O$ with radius $1$ and $\angle{AOQ} = x$, $0 < x < \dfrac{\pi}{2}$

we know that

$$OA = 1$$

so

$$\tan(x) = \dfrac{QA}{1}, \quad QA = \tan(x)$$

calculating area $\Delta OAP$

$$\text{area} = \dfrac{1}{2}ab\sin(x) = \dfrac{1}{2}(1)(1)\sin(x) = \dfrac{1}{2}\sin(x) \\ \; \\ \Delta \text{OAP} = \dfrac{1}{2}\sin(x)$$

calculating area sector $OAP$

$$\text{area} = \dfrac{1}{2}r^2x = \dfrac{1}{2}x \\ \; \\ \text{OAP} = \dfrac{1}{2}x$$

calculating area $\Delta OAQ$

$$\text{area} = \dfrac{1}{2}bh = \dfrac{1}{2}(1)(\tan(x)) = \dfrac{1}{2}\tan(x) \\ \; \\ \Delta\text{OAQ} = \dfrac{1}{2}\tan(x)$$

since $\Delta OAP < OAP < \Delta OAQ$

$$\dfrac{1}{2}\sin(x) < \dfrac{1}{2}x < \dfrac{1}{2}\tan(x) \\ \; \\ \sin(x) < x < \tan(x)$$

divide all parts by $\sin(x)$

$$1 < \dfrac{x}{\sin(x)} < \dfrac{1}{\cos(x)}$$

take the reciprical of all parts

$$1 > \dfrac{\sin(x)}{x} > \cos(x), \quad 0 < x < \dfrac{\pi}{2}$$

since

$$\lim_{x \to 0} \cos(x) = 1$$

this now follows the Pinching Theorem and therefore

$$\boxed{\lim_{x \to 0} \dfrac{\sin(x)}{x} = 1}$$