How to prove that $\lim_{x \to 0} \sin(x) = 0$ using the epsilon-delta definition? How do I prove that 
     $\lim_{x \to 0} \sin(x) = 0$
using the episilon-delta definition of a limit?
Do I have to divide the domain of $x$ into 4 cases for each quadrant?
Update:
Based on the input from @vadim123,
For any $\epsilon>0$
    if there exists some $\delta>0$ satisfying $|x-0|<\delta$, then
we have to show that $|\sin\ x-0|<\epsilon$
Choose $\delta=\epsilon$.
By transitivity of the $<$ and $\le$ relations, 
$|\sin\ x |\le|x|<\epsilon$.
we have now shown that $|\sin\ x| < \epsilon$.
 A: Hint: $|\sin x|<|x|$, for all $x$.
A: For fun, let's make a geometric proof.
Let $O$ be the origin. Draw a unit circle $C$ centered at the origin. Recall that $\sin \theta$ can be defined as the $y$ coordinate of the point on the unit circle lying $\theta$ units counter-clockwise from the positive $x$-axis.
Suppose $\epsilon > 0$. Our goal is to find $\delta > 0$ such that every $\theta$ such that $0 < |\theta| < \delta$ implies $|\sin \theta| < \epsilon$.
Consider the horizontal lines $\ell$ and $\ell'$ defined by $y = \epsilon$ and $y = -\epsilon$. Every point that is on the unit circle whose $y$ coordinate is between $-\epsilon$ and $\epsilon$ lies between these two lines.
If $\epsilon < 1$, $\ell$ intersects the circle in two points.
We may assume $\epsilon < 1$. (do you understand why?)
Of these two points, let $Q$ be the point with positive $x$ coordinate. Similarly, let $Q'$ be the point where $C$ meets $\ell'$ with positive $x$ coordinate.
Note that every point on the unit circle that has positive $x$ coordinate and has $y$ coordinate satisfying $|y| < \epsilon$ lies within the angle $\angle QOQ'$.
Let $\delta$ be the measure of $\angle QOP$. All of the above shows that if $|\theta| < \delta$, we have $|\sin \theta| < \epsilon$.
