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The problem:

Defining $\sin x$ as the leg $b$ of a right triangle with $\angle B=x$ (in radians) and hypotenuse $1$, prove that $$\lim_{x\to 0}\frac{\sin x}x=1$$

(The motivation is to find the derivative of $\sin x$ in a elementary, "pre-Taylor" and "pre-series" context).

I have seen many times a proof that it is based in the fact that the length of the arc $x$ satisfies $$\sin x<x<\tan x$$ or sometimes $$\sin x<x<\sin x+1-\cos x$$ The lower bound is clear because the $\sin x$ is the length of a straight segment and $x$ is the length of a curved segment with the same endpoints. But I find that the upper bound is based on this intuitive fact:

If $F$ and $G$ are two convex subsets of $\Bbb R^2$ and $F\subset G$, then $|\partial F|<|\partial G|$.

($|\partial F|$ is the length of the boundary of $F$).

But I haven't ever seen a proof of that. I tried it myself but I get stuck trying to bound $$\int_s^t\sqrt{1+y'(u)^2}du$$ provided for example that $y''$ is negative and $y(s)=y(t)$. I realize that $y'$ can't be bounded (note for example $y=\sqrt{1-x^2}$, $-1\le x\le 1$).

Question

Is it possible to justify the derivative of $\sin x$ or the inequality about the lengths of bounds? Is there any calculus text with this approach (or similar) to define trigonometical functions?

ajotatxe
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    have you seen the diagram here: https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 – Calvin Khor Apr 25 '19 at 14:35
  • @CalvinKhor Yes, of course, but not the proof using areas instead of lengths. I wonder if it is true that $$\int_s^t\sqrt{1+y_1'(u)^2}du\le\int_s^t\sqrt{1+y_2'(u)^2}du$$ given that $$y_1(u)\le y_2(u),u\in[s,t]$$ $y_1''<0$ and $y_1(t)=y_2(t)=y_1(s)=y_2(s)$. – ajotatxe Apr 25 '19 at 15:10

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The first inequality is easier to justify by comparing areas(How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?), but I suppose the more interesting question is the inequality

$$\text{length}(y_1) \leq\text{length}{(y_2)}$$ under the assumptions that the graphs $y_i$ are concave, $y_1\le y_2$, and $y_1=y_2$ at the endpoints. You can prove it first for polygonal graphs, and then approximate the continuous graph by a sufficiently detailed polygonal graph.

For polygonal paths, you can proceed as follows. The result is obvious(triangle inequality) if $y_1$ is flat. Otherwise, extend the 2nd segment of $y_1$ backwards until you hit $y_2$ at say $X$. The resulting path obtained from going along $y_1$ until $X$, then switching to $y_1$ is strictly longer than $y_1$, by triangle inequality a large number of times. Now we can ignore the curves up to the point $X$, so we have reduced the number $n$ of segments of $y_1$. The result follows from induction in $n$.

A visual aid -

A visual aid

Calvin Khor
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