A doubt on a proof of $\lim \frac{\sin x}{x}$ as $x\to 0$ provided in Simmons's Calculus with Analytic Geometry I'm having difficulty understanding a proof of $\lim_{x\to0}\frac{\sin{x}}{x}=1$ provided in Simmons's Calculus With Analytic Geometry, pg. 72. The proof goes as follows:
Let $P$ and $Q$ be two nearby points on a unit circle, and let $\overline{PQ}$ and $\widehat{PQ}$ denote the lengths of the chord and the arc connecting these points. Then the ratio of the chord length to the arc length evidently approaches 1 as the two points move together:
$\frac{\text{chord lenght}\overline{PQ}}{\text{arc lenght}\widehat{PQ}}\to1$ as $\widehat{PQ}\to0$
With the notion in the figure, this geometric statement is equivalent to
$\frac{2\sin{\theta}}{2\theta}=\frac{\sin{\theta}}{\theta}\to1$ as $\theta\to0$


My doubt is, doesn't this proof simply that $\lim_{x\to0}\frac{\sin{x}}{x}=\frac{0}{0}$? I mean, sure the ratio of $\text{chord lenght}\;\overline{PQ}$ to $\text{arc lenght}\;\widehat{PQ}$ approaches 1 as $\theta$ approaches $0$, but that's because both the numerator and the denominator approach the samue value, which is $0$. How is it any different from saying $\lim_{x\to0}\frac{\sin{x}}{x}=1$ because both $\sin{x}$ and $x$ approach the samue value $0$ as $x\to0$? 

One can interpret the proof in a more abstract, intuitive way. As arc length decreases, its "curviness" disappears (i.e. it becomes a straight line). We can say this as $\sin(\theta) \to \theta $ as $ \theta \to 0$. Taking this into account:
$$\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = \lim_{\theta \to 0} \frac{\theta}{\theta} = \lim_{\theta \to 0} 1 = 1$$
