Strict proof of infinitesimal equivalency between $\sin{x}$ and $x$

Question

When I was teaching infinitesimal equivalency between $\sin(x)$ and $x$ ($x\rightarrow0$) for Calculus, I realized that it was not very easy to have a pure elementary proof for it without using the area formula for sectors. Indeed, the area formula for sectors itself can be derived by this equivalency. It seems that we have a circular argument here. A very smart student suggested that we can simply use the fact that among all shapes with a fixed perimeter the disk has the largest area. She further gave a proof sketch for this statement by symmetry and the fact that when one makes two edges of a triangle orthogonal then its area is increased.

I am wondering if there is any other proof of this equivalency with or without using advanced concepts in Calculus? The difficulty here is that this equivalency is so fundamental in Calculus that many other results (e.g., the derivative of $\sin(x)$ is $\cos(x)$) are based on it so cannot be used to prove it. Meanwhile, elementary mathematics is quite unable to deal with curves.

Answer 1

I'm posting an answer to summarize the rather long comment thread before it likely gets moved to chat. In short, the problem in the question reduced to the fact that most classical proofs that the area of the unit circle is $\pi$ implicitly use $\theta \leq \tan \theta$ for $\theta \in (0,\pi/2)$. Then, the typical proof that $\theta \leq \tan \theta$ uses that the area of the unit circle is $\pi$, which is arguably circular.

For example:

1. Archimedes's proof essentially says "Since the perimeter of the polygon is greater than the circumference of the inscribed circle [...]"
2. The usual proof of partitioning the circle into circular sectors and rearranging into a parallelogram/rectangle often just claims the length is $\pi$ by fiat (e.g. wikipedia's entry on the area of the circle does this) without mentioning it's actually $\pi \frac{\sin\delta_n}{\delta_n}$ for an appropriate $\delta_n$.

---

The problem with a modern, formal proof is "what do we even mean by the perimeter of the circle?" To answer this, we need some calculus.

In short, when it exists, we define the arc-length of the curve $\{(t,f(t)) : t \in [a,b]\}$ to be $$\int_a^b\sqrt{1 + (f'(t))^2}\,dt$$ (alternatively, we could define arc-length differently and then prove this as a theorem; all that matters is this formula gives something that extends polygonal "perimeter" in a natural way and it is sufficiently general for our purposes in this problem)

To consider the unit circle $x^2 + y^2 = 1$, we use $y = f(x) = \sqrt{1 - x^2}$. Then the circumference of the circle can be taken as the definition of $2\pi$, so that $$\begin{align*}\pi &:= \int_{-1}^1\sqrt{1 + (f'(x))^2}\,dx \\ &= 2\int_0^1\frac1{\sqrt{1-x^2}}\,dx\end{align*}$$

The area of the circle, $A$, is given by $$\begin{align*} A &= 2\int_{-1}^1 f(x)\,dx \\ &= 4\int_0^1 \sqrt{1-x^2}\,dx \\ &= 2\int_0^1 \sqrt{1-x^2}\,dx + 2\int_0^1 (x)'\sqrt{1-x^2}\,dx \\ &= 2\int_0^1 \sqrt{1-x^2}\,dx - 2\int_0^1 x\left(\sqrt{1-x^2}\right)'\,dx \\ &= 2\int_0^1 \sqrt{1-x^2}\,dx + 2\int_0^1 \frac{x^2}{\sqrt{1-x^2}}\,dx \\ &= 2\int_0^1 \left(\sqrt{1-x^2} + \frac{x^2}{\sqrt{1-x^2}}\right)\,dx \\ &= 2\int_0^1 \frac1{\sqrt{1-x^2}}\,dx \\ &= \pi \end{align*}$$

Note that this also proves that $\pi < \infty$ and therefore the corresponding arc-length integral exists (which possibly wasn't obvious from the original definition) as $A \leq 4$ since the circle sits inside the square $[-1,1]^2$

Answer 2

This is a way that I discovered it myself, when I was back in Grade 7.

Take a unit circle($r=1$). Inscribe a cyclic regular polygon in it( it's vertices must lie on the circle). The circle will act as the circumcircle of the polygon. It's well known that an $n$ sided regular polygon can be divided into $n$ isoceles triangles, like this:-

Let $\angle CAD=2x=\frac{2π}{n}$. By basic trigonometry, we can say that $CD=2×\sin x$. Thus perimeter of the polygon=$n×(2\sin x)=2n\sin x$.

Now, as $x→0, 2n \sin x→2π$

But $2π=2xn$, so $2n \sin x→2xn$ and thus $\sin x→x$.