calculus - Is the usual proof of $lim_{xrightarrow 0}frac {sin(x)}{x} = 1$ an honest proof?

In a lot of textbooks on Calculus a proof that $\lim_{x\rightarrow 0}\frac {\sin(x)}{x} = 1$ is the following:

Comparing the areas of triangles $ABC, ABD$ and circular sector, you get:
$$
\sin(x)$$
from what you have:
$$
\cos(x)<\frac {\sin(x)}{x}<1
$$

from which immidiately follows that $\lim_{x\rightarrow 0}\frac {\sin(x)}{x} = 1$

Don't you think that this kind of proof is not honest. I mean, in the proof we use the fact that the area of sector is $\frac12xr^2$. This formula comes from integrating (an "infinite" sum of "infinitesimal" triangles' areas). So, roughly speaking, we somehow implicitly use that the length of "infinitesimal" chord $BC$ is equal to the length of "infinitesimal" arch $rx$, which is equivalent that $\lim_{x\rightarrow 0}\frac {\sin(x)}{x} = 1$.

Do I miss something?

Answer

It is a perfectly honest proof, but it is based on few assumptions about areas bounded by plane curves and the definitions of trigonometric functions. The main assumption is that a sector of a circle has an area. The proof of this assumption requires real analysis/calculus but it does not require the limit $\lim_{x\to 0}\dfrac{\sin x}{x}=1$. Once this assumption is established, the next step is to define the number $\pi$ as area of unit circle (circle of radius $1$).

Next we consider the same figure given in question. We need to define a suitable measurement of angles. This can be done in many ways, and one of the ways is to define the measure of $\angle CAB$ as twice the area of sector $CAB$. Note that for this definition to work it is essential that radius of sector is $1$ (which is the case here).

Further we define the functions $\sin x,\cos x$ in the following manner. Let $A$ be origin of the coordinate axes and $AB$ represent positive $x$-axis. Also let the measure of $\angle CAB$ be $x$ (so that area of sector $CAB$ is $x/2$). Then we define the coordinates of point $C$ as $(\cos x,\sin x)$.

Now that we know these assumptions, the proof presented in the question is valid and honest. This is one of the easiest routes to a proper theory of trigonometric functions. The real challenge however is to show that a sector of a circle has an area.