TITLE: A doubt on a proof of $\lim \frac{\sin x}{x}$ as $x\to 0$ provided in Simmons's Calculus with Analytic Geometry
QUESTION [8 upvotes]: 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 $x\to0$?

REPLY [4 votes]: The questions raised in the last paragraph of your post are best handled by the excellent answer from Ethan Bolker. The following is a feedback on the proof given in your book.

I really can't believe that this is an actual proof given in an actual book. If this is really the case as you say, then this is one of the boldest examples of intellectual dishonesty perpetrated by most calculus textbook authors.
Note that $$\lim_{x\to 0}\frac{\sin x} {x} = 1\Rightarrow \frac{\text{chord length } PQ} {\text{arc length } PQ} \to 1 \text{ as arc length }PQ\to 0$$ and not the other way round as the proof in your book seems to suggest. For the case of a circle, the ratio of chord length to arc length tends to $1$ but this is not as self evident as Simmons writes, rather it depends on the limit in question here. The proof presented in the book is just plain hand-waving.
The proof for the limit can be easily given using areas of sectors instead of lengths of arcs because the analysis of arc length is significantly more cumbersome compared to that of areas. The standard proof is the one mentioned by user robjohn in this answer.
When using arc length we have two options:

Establish that the ratio of chord length to arc length tends to $1$ as the arc length tends to $0$ for a general class of sufficiently well behaved curves (and circle is one such curve) and then argue as in your book. This can be done by having a rigorous analysis of length of arc of a general curve and using some theorems from integral calculus.
Treat the case of circle on special basis and establish the inequality $\sin x<x<\tan x$ for $x\in(0,\pi/2)$ using length of arc of a circle and then apply squeeze theorem.