calculus - Is the proof of $lim_{thetato 0} frac{sin theta}{theta}=1$ in some high school textbooks circular?

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

$$\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1$$