TITLE: Is the proof of $\lim_{\theta\to 0} \frac{\sin \theta}{\theta}=1$ in some high school textbooks circular?
QUESTION [16 upvotes]: I was taught the following proof in high school.

By constructing triangles with $0<\theta<\pi/2$ and a circle with radius $r$ and by comparing the areas, we have
$$\frac{1}{2}r^2\sin\theta\cos\theta \le \frac{1}{2}r^2\theta\le\frac{1}{2}r^2\tan\theta$$
Hence
$$\cos\theta\le\frac{\theta}{\sin\theta}\le\frac{1}{\cos\theta}$$
Then by squeeze theorem, we have the result.
My question is, the middle term in the above inequality comes from the fact that the area of the circle is $\pi r^2$, which in my textbooks, is later proved by integration. But the integration requires results in calculus which comes from the fact that
$$\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1$$

REPLY [3 votes]: This has been discussed earlier and I will post link if I find it. To simplify the proof we can just work with circle of unit radius. And instead of dealing with arc lengths we deal with area of sectors. This avoids the problem of dealing with comparison of arc lengths which requires notions of convexity of curves (like done in Jack d'Aurizio's answer here).
The following theorem holds:

If $L$ is the length of an arc of unit circle and $A$ the area of corresponding sector then $L=2A$.

The proof follows not via the use of circular functions as many believe but simply via integration by parts. Note that we have $$\int_{a} ^{1}\sqrt{1-x^{2}}\,dx=-\frac{a\sqrt{1-a^{2}}}{2}+\frac{1}{2}\int_{a}^{1}\frac{dx}{\sqrt{1-x^{2}}}$$ or $$\int_{a}^{1}\frac{dx}{\sqrt{1-x^{2}}}=2\left(\frac{a\sqrt{1-a^{2}}}{2}+\int_{a}^{1}\sqrt{1-x^{2}}\,dx\right)$$ which is the desired relation $L=2A$ for an arc with end points $(1,0)$ and $(a, \sqrt{1-a^{2}})$ (with $a\in(0,1)$) of the unit circle. Once we have this relation we immediately get the following inequality (as mentioned in question) via comparison of areas: $$\sin\theta<\theta<\tan\theta$$ for all $\theta\in(0,\pi/2)$.

Update: If one translates the geometrical definition into analysis we get the the following link between $x$ and $\sin x$ which is valid for $x\in(-\pi/2,\pi/2)$: $$x=\int_{0}^{\sin x} \frac{dt} {\sqrt{1-t^{2}}}$$ and dealing with positive values of $x$ we see that the integrand in the integral on right of above equation lies between $1$ and $1/\sqrt{1-\sin^{2}x}$ and thus we get the inequalities $$\sin x< x< \frac{\sin x} {\sqrt{1-\sin^{2}x}}=\tan x$$ so all that argument from geometry about arc-length of convex curves is not needed. The traditional proof avoids usage of direct formula for length of arc of a circle in terms of integrals and the argument becomes unnecessarily complex. This is done because the proof is presented at a stage when one is learning limits and is totally unaware of the concept of an integral.