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
$$\lim_{\theta->0}\sin\theta=0.$$
Best Answer
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: $$\,=-0=0.$$