Non-Circular Proof of $\lim_{x \to 0} \frac{\sin x}{x} = 1$ I'm looking for a convincing proof, using first principles, that $$\lim_{x \to 0}\frac{\sin x}{x} = 1$$ (Please use ordinary unit circle definitions of trigonometric functions.)
It occurred to me that the classic proof, which compares three areas, uses the formula ${1\over 2}r^2\theta$ for the area of a circular sector of angle $\theta$, which in turn assumes the area of a circle is $\pi r^2$. But this fact is almost always proven in texts using an integral, which ends up using the derivatives of $\sin$ and $\cos$, and we're back to that limit again.
So I need a non-circular proof that doesn't rely on playing definition games ("let $\sin$ be the following power series..."). The answer to this question is definitely playing definition games.
Sorry for the pun.
 A: I don't see anything circular in comparing areas to get the inequality $\sin x < x < \tan x$ for $0 < x < \pi/2$. However we need to be very cautious in defining the symbols $\sin x, \tan x$ properly given $x$ a real number.
The approach based on areas goes like this. Using the concept of definite integrals it can be proven that a sector of a circle has an area. This does not require anything beyond the continuity of the function $\sqrt{1 - x^{2}}$ in interval $[0, 1]$. In particular justification of the area of a circle is not dependent on the definition of trigonometric functions and $\pi$.
Next consider a unit circle with origin $O$ as center and let $A$ be the point $(1, 0)$. Let $P$ be any point on the circle. For our purposes it is sufficient to consider $P$ to be in first quadrant. Let the area of sector $AOP$ be $y$ so that $y > 0$. Also let $x = 2y$ and then by definition the point $P$ is $(\cos x, \sin x)$. This is the usual definition of trigonometric functions as studied at the age of 15 years or so.
Note that some textbooks base the definition of $\sin x, \cos x$ on the basis of length of arc $AP$ which is $x$. The definition is equivalent to the one based on areas of sectors, but comparing areas of figures is simpler than comparing the length of arcs (at least in this context). Consider the tangent $AT$ to unit circle at point $A$ such that $OPT$ is a line segment. Also let $PB$ be a perpendicular to $OA$ and $B$ is the foot of this perpendicular. Now it is easy to show that $$\text {area of }\Delta AOP < \text{ area of sector }AOP < \text{ area of }\Delta AOT$$ (because each region is contained in the next). However it is very difficult to compare the length of arc $AP$ with the length of line segments $PB$ and $AT$ (because there is no containment here).
The above inequality leads to $$\sin x < x < \tan x$$ from which we get $\sin x \to 0$ as $x \to 0$ and then $\cos x = \sqrt{1 - \sin^{2}x} \to 1$. Further the inequality is equivalent to $$\cos x < \frac{\sin x}{x} < 1$$ and hence $(\sin x)/x \to 1$ as $x \to 0$.

Update: It appears from OP's comments that the relation between length of an arc of a circle and area of corresponding sector is something which can't be proven without using any analytic properties of circular functions. However this is not the case.
Let $P = (a, b)$ be a point on unit circle $x^{2} + y^{2} = 1$ and let $A = (1, 0)$. For simplicity let's consider $P$ in first quadrant so that $a, b$ are positive. Then the length of arc $AP$ is given by $$L = \int_{a}^{1}\sqrt{1 + y'^{2}}\,dx = \int_{a}^{1}\frac{dx}{\sqrt{1 - x^{2}}}$$ The area of the sector $AOP$ is given by $$A = \frac{ab}{2} + \int_{a}^{1}\sqrt{1 - x^{2}}\,dx$$ We need to prove that $L = 2A$. We will do this using the fact that $b = \sqrt{1 - a^{2}}$ and using integration by parts.
We have
\begin{align}
\int\sqrt{1 - x^{2}}\,dx &= x\sqrt{1 - x^{2}} - \int x\cdot\frac{-x}{\sqrt{1 - x^{2}}}\,dx\notag\\
&= x\sqrt{1 - x^{2}} - \int \frac{1 - x^{2} - 1}{\sqrt{1 - x^{2}}}\,dx\notag\\
&= x\sqrt{1 - x^{2}} - \int \sqrt{1 - x^{2}}\,dx + \int \frac{1}{\sqrt{1 - x^{2}}}\,dx\notag\\
\Rightarrow \int\sqrt{1 - x^{2}}\,dx &= \frac{x\sqrt{1 - x^{2}}}{2} + \frac{1}{2}\int \frac{dx}{\sqrt{1 - x^{2}}}\notag\\
\end{align}
Hence $$\int_{a}^{1}\sqrt{1 - x^{2}}\,dx = - \frac{a\sqrt{1 - a^{2}}}{2} + \frac{1}{2}\int_{a}^{1}\frac{dx}{\sqrt{1 - x^{2}}}$$ or $$\int_{a}^{1}\frac{dx}{\sqrt{1 - x^{2}}} = 2\left(\frac{ab}{2} + \int_{a}^{1}\sqrt{1 - x^{2}}\,dx\right)$$ or $L = 2A$ which was to be proved.
Contrast the above proof of relation between length and area with the following totally non-rigorous proof. Let the length of arc $AP$ be $L$. Then the angle subtended by it at the center is also $L$ (definition of radian measure). Divide this angle into $n$ parts of measure $L/n$ each and then the area of sector $AOP$ is sum of areas of these $n$ sectors. If $n$ is large then area of each of these $n$ sectors can be approximated by area of the corresponding triangles and this area is $$\frac{1}{2}\sin (L/n)$$ so that the area of the whole sector $AOP$ is $(n/2)\sin(L/n)$. As $n \to \infty$ this becomes $L/2$ and here we need the analytic property of $\sin x$ namely $(\sin x)/x \to 1$ as $x \to 0$. Therefore area can't be the basis of a proof of this limit. This is perhaps the reason that proofs for limit formula $(\sin x)/x \to 1$ looks circular.
A proper proof can't be done without integrals as I have shown above. Hence the proof that $(\sin x)/x \to 1$ depends upon Riemann integration and definition of $\sin x, \cos x$ as inverses to the integrals. This is same as $e^{x}$ is defined as inverse to integral of $1/x$.
Also see my another answer to a similar question.
A: You show that $x<\tan x$ when $0<x<\pi/2$ by taking the path:
$$(\cos 2x,\sin 2x),(1,\tan x),(1,0)$$
This path, aside from the endpoints, is outside the unit disk. The two segments are tangent to the circle. There is a theorem which says that therefore it has to be longer than the shortest arc on the boundary of the unit disk. I asked a question about this a while back - it is a result of the Hahn Banach theorem.
So this path, of length $2\tan x$, is greater than or equal to the path on the circle, $2x$.
(That $\sin x\leq x$ is due to a simpler rule - the shortest path from a point to a line, is the perpendicular from the point to the line.)
A: You can also note that (for small, positive $x$) we have $$\sin(x)<x<\tan(x)$$then divide through by $\sin(x)$ and squeeze. This gives the limit of $\frac{x}{\sin x}$.
A: $\sin x$ is easy to define, but not $x$! You won't find a more elementary
rigorous definition of the $\sin$ function other that is the inverse of
the $\sin^{-1}$ function, where $\sin^{-1} y$ is defined for not too large
$y$ as the length of the arc in the circle between $(1, 0)$ and
$(\sqrt{(1-y^2)}, y)$. This is just the completely elementary definition
with the units for the angle actually defined.  It is not elementary
enough for a first course since it uses inverse functions.  The existence
of the arc length can be hand-waved and the coordinates not stated in a
first course.
Rigorously defining arc lengths needs a limit operation.  Fairly
elementary calculus gives the formula for the arc length of a circle:
$$\sin^{-1} y = \int ^y _0 \frac{dt}{\sqrt{(1-t^2)}}.$$  Note that
this is not circular; it is the definition of $\sin^{-1}$ and thus of
$\sin$.
The value of the limit is obvious from this formula using elementary
calculus.  Note that I haven't done anything complicated like evaluating
the integral.  Other properties of the $\sin$ function can be derived,
not so easily, from this definition.  $\pi$ is by definition the length
of a half-circle, and thus twice the value of this integral from $0$ to
$1$, whatever that may be.  Since the integral becomes improper at
$y = 1$, better define $\pi$ using half of the first quadrant instead
of all of it to minimize the calculus.
This is easier, at least for analysts, to understand after studying
elliptic functions and elliptic integrals.  Change $t^2$ to $t^3$ in
the integral and the integral becomes elliptic. There is no geometry of
the circle or arc length to help or confuse you in understanding this
integral. The inverse of the elliptic integral is an elliptic
(doubly periodic) function. This is much less easy to see and prove
rigorously from the integral than that $sin$ is singly periodic. Other properties of elliptic functions are even harder to see and prove
starting from the integral.
