Evaluate $\lim_{x \to 0} \dfrac{\sin(2x)}{8x}$ I need to evaluate 
$$\lim_{x \to 0} \dfrac{\sin(2x)}{8x}$$
I'm doing Calculus One course on coursera.org and here is their explanation that doesn't explain anything for me:

Why $\lim_{x \to 0} \dfrac{\sin(x)}{x}=1$ ?

Why do I need to set $y=2x$? And why $y\to 0 as x\to 0$?

Why $\lim_{y \to 0} \dfrac{\sin(y)}{y} \cdot \dfrac{2}{8}$ suddenly became $\lim_{x \to 0} \dfrac{\sin(2x)}{8x}=\dfrac{1}{4}$?
For the first one, why not google it?
Then 
$$\lim_{x\to 0}\frac{\sin 2x}{8x}=\lim_{2x\to 0}\frac{\sin 2x}{4\cdot 2x}=\frac14\lim_{2x\to 0}\frac{\sin 2x}{2x}=\frac14\cdot 1=\frac14$$
$\bf hint:$ if you have you a calculator that can compute $\sin,$ then compute $\frac{\sin x}{x}$ for $x = 0.1, 0.01, 0.001$ and see what happens. look at the graphs of $\sin x$ and $x$ in the window $-1 \le x \le 1, -1 \le y \le 1$ make sure your calculator is in radian mode. happy experimenting.