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 $x\to0$?
I think your problem is deeper than understanding this proof.

What makes calculus difficult is that it's hard (at first) to make sense of the ratio of two small quantities. Just because both "approach the same value $0$" doesn't mean the limit of the quotient is $1$. 

For example
$$
\frac{2x}{x}
$$
clearly approaches $2$, since it's always exactly $2$ when $x$ is not $0$.

Also
$$
\frac{2h+h^2}{h} = 2 + h
$$
approaches $2$ as $h$ approaches $0$, even though it looks like $0/0$ when $h = 0$.
That is in fact just the limit you probably saw when finding the slope of a parabola (differentiating $y=x^2$ when $x=1$).

The moral of the story is that you have to pay attention to the relative rates at which the numerator and denominator approach $0$. In the question you ask, the rates are the same and the limit is $1$. 

You can take comfort in the fact that this was a mathematical and philosophical difficulty when Newton and Leibniz were inventing calculus.

Note: this in an informal discussion. The formal definition of limits avoids words like "approaches". And @Bernard 's comment about the geometric argument for $\sin(x)/x$ correctly notes the informality there too.
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:

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