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.
The area of $\triangle ABC$ is $\frac{1}{2}\sin(x)$.  The area of the colored wedge is $\frac{1}{2}x$, and the area of $\triangle ABD$ is $\frac{1}{2}\tan(x)$.  By inclusion, we get
$$
\frac{1}{2}\tan(x)\ge\frac{1}{2}x\ge\frac{1}{2}\sin(x)\tag{1}
$$
Dividing $(1)$ by $\frac{1}{2}\sin(x)$ and taking reciprocals, we get
$$
\cos(x)\le\frac{\sin(x)}{x}\le1\tag{2}
$$
Since $\frac{\sin(x)}{x}$ and $\cos(x)$ are even functions, $(2)$ is valid for any non-zero $x$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.  Furthermore, since $\cos(x)$ is continuous near $0$ and $\cos(0) = 1$, we get that
$$
\lim_{x\to0}\frac{\sin(x)}{x}=1\tag{3}
$$
Also, dividing $(2)$ by $\cos(x)$, we get that
$$
1\le\frac{\tan(x)}{x}\le\sec(x)\tag{4}
$$
Since $\sec(x)$ is continuous near $0$ and $\sec(0) = 1$, we get that
$$
\lim_{x\to0}\frac{\tan(x)}{x}=1\tag{5}
$$
This is not a rigorous proof, but is instead an intuitive argument. Consider the graph of the sine function and in particular consider the origin $(0,0)$ and some arbitrary point $(x,\sin(x))$ a little bit to the right of it. Connect a secant line between the two points like so:
Now consider the average rate of change of the sine function on that interval $[0,x]$, or alternatively the slope $m$ of that secant line, namely:
$$m(x)=\frac{\sin(x)-\sin(0)}{x-0}$$
But clearly this is just
$$m(x)=\frac{\sin(x)}{x}$$
But then &quot;clearly&quot; (again, this is just for intuitive purposes), as $x\to{0}$, the slope of this secant line approaches the slope of the tangent line to the graph at $x=0$ (i.e. it approaches whatever $\sin'(0)$ should be). Looking at the graph then, it shouldn't feel too unreasonable that this limiting procedure results in a $45$-degree diagonal tangent, hence a slope of $1$.