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 $$\lim_{\theta->0}\sin\theta=0.$$
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$\begingroup$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: $$\,=-0=0.$$
$\endgroup$ 0 $\begingroup$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 $$ \lim_{x\to0}\sin x=0 $$
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