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.

Let $\sin(x)$ is defined as solution of $\frac{d^2}{dx^2}\textrm{f}(x)=-\textrm{f}(x)$ with $\mathrm f(0)=0,\,\frac{d}{dx}\mathrm f(0)=C$ initial conditions, so exact solution is $\mathrm f(x)=C\cdot\sin(x)$.
 Define second derivative as 
$$
\begin{align*}
\frac{d^2}{dx^2}\textrm{f}(x)=\lim_{\Delta x\to 0}{\frac{\frac{\mathrm f(x)-\mathrm f(x-\Delta x)}{\Delta x}-\frac{\mathrm f(x-\Delta x)-\mathrm f(x-2\cdot\Delta x)}{\Delta x}}{\Delta x}}&=\\=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)-2\cdot \mathrm f(x-\Delta x)+\mathrm f(x-2\cdot\Delta x)}{\Delta x^2}}
\end{align*}
$$
we can easy check this limit for any (?) functions. Similarly, we can define the first derivative for right, middle and left points:
$$
\frac{d}{dx}\textrm{f}(x)\left\{
\begin{aligned}
&=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)-\mathrm f(x-\Delta x)}{\Delta x}}
\\ 
&=\lim_{\Delta x\to 0}{\frac{\mathrm f(x-\Delta x)-\mathrm f(x-2\cdot\Delta x)}{\Delta x}}\\
&=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)-\mathrm f(x-2\cdot\Delta x)}{2\cdot\Delta x}}
\end{aligned}
\right.
$$
 Let's use the finite elements method assuming $Td=\Delta x,\,y_n=\mathrm f(x),\,y_{n-1}=\mathrm f(x-\Delta x),\,y_{n-2}=\mathrm f(x-2\cdot \Delta x)$
Override differential equation as 
$$
\frac{y_n-2\cdot y_{n-1}+y_{n-2}}{Td^2}=-y_n
$$
Now solve this implicit equation for $y_n$ to obtain explicit recurrence relation:
$$
y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2}
$$
Using arbitrarily small but non-zero quantity Td we can plot exponentially decaying sampled sine function (because the poles are inside the unit circle of the transfer function corresponding to the given recurrence relation).
Similarly we write three systems for the initial conditions:
$$
\left\{
\begin{aligned}
&y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2}
\\ 
&C=\frac{y_n-y_{n-1}}{Td}
\end{aligned}\right.
$$
$$
\left\{
\begin{aligned}
&y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2}
\\ 
&C=\frac{y_{n-1}-y_{n-2}}{Td}
\end{aligned}\right.
$$
$$
\left\{
\begin{aligned}
&y_n = \frac{2\cdot y_{n-1}-y_{n-2}}{1+Td^2}
\\ 
&C=\frac{y_n-y_{n-2}}{2\cdot Td}
\end{aligned}\right.
$$
Solve this sequence of equations for $y_{n-1}$ and $y_{n-2}$:
$$
\left\{
\begin{aligned}
&y_{n-1} = -C\cdot Td + y_{n}\\ 
&y_{n-2}=-2\cdot C\cdot Td + y_{n}\cdot\left(1-Td^2\right)\
\end{aligned}\right.
$$
$$
\left\{
\begin{aligned}
&y_{n-1} = -C\cdot Td + y_{n}\cdot\left(1+Td^2\right)\\ 
&y_{n-2}=-2\cdot C\cdot Td + y_{n}\cdot\left(1+Td^2\right)\
\end{aligned}\right.
$$
$$
\left\{
\begin{aligned}
&y_{n-1} = -C\cdot Td + y_{n}\cdot\left(1+\frac{Td^2}{2}\right)\\ 
&y_{n-2}=-2\cdot C\cdot Td + y_{n}\
\end{aligned}\right.
$$
At zero point $y_n=\mathrm f(0)=0$ and we can see linear dependence: 
$$
\begin{aligned}
 &y_{n-1} = -C\cdot Td\\
 &y_{n-2}=-2\cdot C\cdot Td 
\end{aligned}
$$
 for all three solutions. Replace back:
 $$
\begin{array}{l}
 \mathrm f(0)&=0\\
 \mathrm f(0-\Delta x) &= -C\cdot \Delta x\\
 \mathrm f(0-2\cdot \Delta x) &= -2\cdot C\cdot \Delta x
\end{array}
$$
So all three $\frac{d}{dx}\mathrm f(0)$ limits is equal to $C$ at $x=0$ and in accordance with $\mathrm f(x)=C\cdot\sin(x)$ by definition we can write 
$$
\lim_{\Delta x\to 0}{\frac{\mathrm f(0)-\mathrm f(0-\Delta x)}{\Delta x}}=\lim_{\Delta x\to 0}{\frac{0-(-C \cdot \Delta x)}{\Delta x}}=C
$$ 
Thus
$$
\lim_{\Delta x\to 0}{\frac{\sin(0)-C\cdot\sin(0-\Delta x)}{\Delta x}}=\lim_{\Delta x\to 0}{\frac{C\cdot\sin(\Delta x)}{\Delta x}}=C\cdot\lim_{\Delta x\to 0}{\frac{\sin(\Delta x)}{\Delta x}}=C
$$ 
and $\lim_{\Delta x\to 0}{\frac{\sin(\Delta x)}{\Delta x}}=1$
