Limit of $\sin(x)$ as $x$ approaches zero from the left I'm trying to find a proof that:
$$ \lim_{x\to0^-}\sin(x)=0$$
I'd like to be able to do the proof without reference to advanced theorems (mean value theorem, series, etc). I have a geometric approach for finding the limit from the right, but I need similar help when approaching zero from the left.
Thanks.
Update: I am about to prove that the sine is continuous at any value $a$, but I first need to prove that 
$$\lim_{\theta\to0}\sin\theta=0\quad\text{and}\quad \lim_{\theta\to0}\cos\theta=1.$$ 
I've already shown that $f$ is continuous at $a$ if 
$$\lim_{h\to 0}f(a+h)=f(a),$$ 
so then I can show 
$$\lim_{h\to0}\sin(a+h)=\sin(a),$$ 
which implies that the sine is continuous at any $a$. But to do that last step, I need 
$$\lim_{\theta\to0}\sin\theta=0 \quad\text{and}\quad \lim_{\theta\to0}\cos\theta=1.$$ 
Thus, I need to prove each of these without using continuity.
I've shown that $\lim_{\theta\to0^+}\sin\theta=0$ using the following image:

The area of the triangle is $\frac12r^2\sin\theta$ and the area of the sector is $\frac12r^2\theta$, so,
$$0\le\frac12r^2\sin\theta\le\frac12r^2\theta,$$
which simplifies to:
$$0\le \sin\theta\le \theta$$
By the squeeze theorem, this makes $\lim_{\theta\to0+}\sin\theta=0$. Now I need a proof that $\lim_{\theta\to0^-}\sin\theta=0$.
Update: Due to all the nice help I received, it turns out that if $0\le\theta\le\pi/2$, then
$$\sin\theta\le\theta$$
which, which because $\sin\theta$ and $\theta$ are both positive on $0\le\theta\le\pi/2$, is equivalent to
$$|\,\sin\theta\,|<|\,\theta\,|.$$
Secondly, if $-\pi/2\le\theta\le0$, then $0\le-\theta\le\pi/2$. Hence, we can substitute $-\theta$ in the last inequality, which leads to:
$$\begin{align*}
|\sin(-\theta)\,|&\le|-\theta\,|\\
|-\sin(\theta)\,|&\le|-\theta\,|\\
|\sin(\theta)\,|&\le|\,\theta\,|
\end{align*}$$
Therefore, if $-\pi/2\le\theta\le\pi/2$, then
$$|\sin(\theta)\,|\le|\,\theta\,|.$$
The last step is due to the fact that $|-x|=|x|$ for all real numbers $x$. This last inequality is equivalent to
$$-|\,\theta\,|\le\sin\theta\le|\,\theta\,|,$$
and by the squeeze theorem, since both ends go to zero as $\theta\to0$, I've shown that
$$\lim_{\theta->0}\sin\theta=0.$$
 A: As egreg suggests in the comments, first reverse the direction of the one-sided limit:
$$\lim_{x\to0^-}\sin(x)=\lim_{x\to0^+}\sin(-x)$$
Then use the fact that the sine function is odd:
$$\,=\lim_{x\to0^+}\left[-\sin(x)\right]$$
Bring the constant factor of $-1$ outside the limit:
$$\,=-\lim_{x\to0^+}\sin(x)$$
Now use your earlier result:
$$\,=-0=0.$$
A: Note that it's not necessary to prove continuity via the limits from the left and from the right. Just prove that, for every $\varepsilon>0$, the solutions of the inequality
$$
|\sin(a+h)-\sin a|<\varepsilon
$$
(in the unknown $h$) form a neighborhood of $0$.
The inequality can be rewritten
$$
\left|2\cos\frac{2a+h}{2}\sin\frac{h}{2}\right|<\varepsilon
$$
Since $|\cos t|\le1$, we are done if we find that the solutions of
$$
\left|\sin\frac{h}{2}\right|<\frac{\varepsilon}{2}\tag{*}
$$
form a neighborhood of $0$.
Saying that, for every $\varepsilon>0$, the solutions of the inequality (*) form a neighborhood of $0$ is the same as stating that the sine function is continuous at $0$.
If you prove with the geometric definition that $\sin x\le x$, for $0\le x<\pi/2$, then you're done, because the symmetry of $\sin x$ also gives
$$
x\le\sin x
$$
for $-\pi/2<x\le0$: just recall that $\sin(-x)=-\sin x$.
So, for $-\pi/2<x<\pi/2$, we have $|\sin x|\le|x|$.

If you want to do this with limits, the inequality 
$$-|x|\le \sin x\le |x|$$
gives immediately that
$$
\lim_{x\to0}\sin x=0
$$
