$\epsilon-\delta$ argument for continuity for $f(x) = \sin(x)$ at $x=0$ Given
$$\lim_{x\to c} f(x) = L$$
We want to show that $\forall \epsilon&gt;0, \exists \delta&gt;0$ such that 
$$|x-c| &lt; \delta \implies |f(x) - L | &lt; \epsilon$$

Our $f(x)=\sin(x)$ and our $c=0$, and $L=0$. So, we suppose 
$|x|&lt;\delta$ and need to show $|\sin(x)|&lt;\epsilon$.

My teacher did this by drawing a picture and showing it holds, but I wanted to see how it would work strictly algebraically.
hint: consider $g(x)=x-sin(x)$. $g'(x)=1-cos(x) \ge 0$ which means that $g(x)$ is increasing and because $g(0)=0$ we conclude that $x - sin (x) \ge 0$
As was already written, it depends on definition of sine.
But if we suppose y-coordinate of point of unit circle as definition,
it is obvious that for $\varepsilon&gt;1$ is any $\delta$ good
and if $0&lt;\varepsilon&lt;1$ then $\delta=\inf{\{t;t&gt;0\wedge \sin{\left(t\right)}&gt;\varepsilon\}}$