In a lot of textbooks on Calculus a proof that limx→0sin(x)x=1 is the following:
Comparing the areas of triangles ABC,ABD and circular sector, you get:
$$
\sin(x)
\cos(x)<\frac {\sin(x)}{x}<1
$$
from which immidiately follows that limx→0sin(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 12xr2. 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 limx→0sin(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 limx→0sinxx=1. Once this assumption is established, the next step is to define the number π 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 ∠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 sinx,cosx in the following manner. Let A be origin of the coordinate axes and AB represent positive x-axis. Also let the measure of ∠CAB be x (so that area of sector CAB is x/2). Then we define the coordinates of point C as (cosx,sinx).
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.
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