How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$? How can one prove the statement
$$\lim_{x\to 0}\frac{\sin x}x=1$$
without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution.
This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statement is enough for proving the continuity of $\sin$, but I can't find out how. Any help is appreciated.

Usual proofs can be circular, but there is a simple way for proving such inequality.
Let $\theta$ be an acute angle and let $O,A,B,C,D,C'$ as in the following diagram:

We may show that:
$$ CD \stackrel{(1)}{ \geq }\;\stackrel{\large\frown}{CB}\; \stackrel{(2)}{\geq } CB\,\stackrel{(3)}{\geq} AB $$
$(1)$: The quadrilateral $OCDC'$ and the circle sector delimited by $O,C,C'$ are two convex sets. Since the circle sector is a subset of the quadrilateral, the perimeter of the circle sector is less than the perimeter of the quadrilateral.
$(2)$: the $CB$ segment is the shortest path between $B$ and $C$.
$(3)$ $CAB$ is a right triangle, hence $CB\geq AB$ by the Pythagorean theorem.
In terms of $\theta$ we get:
$$ \tan\theta \geq \theta \geq 2\sin\frac{\theta}{2} \geq \sin\theta $$
for any $\theta\in\left[0,\frac{\pi}{2}\right)$. Since the involved functions are odd functions the reverse inequality holds over $\left(-\frac{\pi}{2},0\right]$,  and $\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1$ follows by squeezing.

A slightly different approach might be the following one: let us assume $\theta\in\left(0,\frac{\pi}{2}\right)$. By $(2)$ and $(3)$ we have
$$ \theta \geq 2\sin\frac{\theta}{2}\geq \sin\theta $$
hence the sequence $\{a_n\}_{n\geq 0}$ defined by $a_n = 2^n \sin\frac{\theta}{2^n}$ is increasing and bounded by $\theta$. Any increasing and bounded sequence is convergent, and we actually have $\lim_{n\to +\infty}a_n=\theta$ since $\stackrel{\large\frown}{BC}$ is a rectifiable curve and for every $n\geq 1$ the $a_n$ term is the length of a polygonal approximation of $\stackrel{\large\frown}{BC}$ through $2^{n-1}$ equal segments. In particular
$$ \forall \theta\in\left(0,\frac{\pi}{2}\right), \qquad \lim_{n\to +\infty}\frac{\sin\left(\frac{\theta}{2^n}\right)}{\frac{\theta}{2^n}} = 1 $$
and this grants that if the limit $\lim_{x\to 0}\frac{\sin x}{x}$ exists, it is $1$. By $\sin x\leq x$ we get $\limsup_{x\to 0}\frac{\sin x}{x}\leq 1$, hence it is enough to show that $\liminf_{x\to 0}\frac{\sin x}{x}\geq 1$. We already know that for any $x$ close enough to the origin the sequence $\frac{\sin x}{x},\frac{\sin(x/2)}{x/2},\frac{\sin(x/4)}{x/4},\ldots$ is convergent to $1$, hence we are done.
Long story short: $\lim_{x\to 0}\frac{\sin x}{x}=1$ follows from the fact that a circle is a rectifiable curve, and a circle is a rectifiable curve because it is the boundary of a convex, bounded subset of $\mathbb{R}^2$. The convexity of the disk follows from the triangle inequality: a disk is a closed ball for the euclidean distance.

$(1)$ relies on this powerful Lemma:

Lemma. If $A,B$ are convex bounded sets in $\mathbb{R}^2$ and $A\subsetneq B$, the perimeter of $A$ is less than the perimeter of $B$.

Proof: by boundedness and convexity, $\partial A$ and $\partial B$ are rectifiable, with lengths $L(A)=\mu(\partial A),\,L(B)=\mu(\partial B)$. Always by convexity, there is some chord in $B$ that does not meet the interior of $A$ (a tangent to $\partial A$ at a smooth point does the job, for instance). Assume that such chord has endpoints $B_1, B_2 \in \partial B$ and perform a cut along $B_1 B_2$: both the area and the perimeter of $B$ decrease, but $B$ remains a bounded convex set enclosing $A$. Since $A$ can be approximated through a sequence of consecutive cuts, $L(A)<L(B)$ follows.
