Let us recall the classic result. What is the limit of $\frac{\sin x}{x}$ at $x=0$?
Without using calculus, we usually prove that by squeezing with the following geometric argument.
The area of triangle $CAB$ is $\frac{1}{2}\sin x$.
The area of sector $CAB$ is $\frac{x}{2}$.
The area of triangle $EAB$ is $\frac{1}{2}\tan x$.
Thus we have the inequality $\sin x \leq x \leq \tan x$ in an neighborhood around $x = 0$.
Rearranging gives $1\leq \frac{x}{\sin x} \leq \sec x$, but since $\sec x \to 1$ as $x\to 0$, squeezing gives $\lim _{x\to 0} \frac{x}{\sin x} = 1$.
That is essentially saying that the first order term of $\sin x$ is $1$ (while the constant term is zero).
Now the question is, how can we calculate higher order terms again without calculus?
For example, what is $\lim _{x\to 0} \frac{1-\cos x}{x^2}$?
With the abundance of trig identities, there are quite a number of possible approaches. Let us start with a pure algebraic one.
For the lower bound, note that
$1-\cos x \geq \frac{1}{2}(1-\cos ^2 x) = \frac{1}{2} \sin ^2 x$, so
$\frac{1-\cos x}{x^2} \geq \frac{1}{2} \frac{\sin ^2 x}{x^2} \to \frac{1}{2}$.
For the upper bound, we use that $\cos x \geq \sqrt{1-x^2}$ that
$\frac{1-\cos x}{x^2} \leq \frac{1 - \sqrt{1-x^2}}{x^2} \leq \frac{1}{1+\sqrt{1-x^2}} \to \frac{1}{2}$, so squeezing gives the answer.
But wait! Isn't it such a pity if we are dealing with a limit with actual geometric interpretation? After all, $1-\cos x$ is the length of segment $BD$. In that case, allow me to present my 'geometric' approach:
We start with the $\sin \frac{x}{2} \leq \frac{x}{2} \leq \tan \frac{x}{2}$, squaring gives $\sin ^2 \frac{x}{2} \leq \frac{x^2}{4} \leq \tan ^2 \frac{x}{2}$.
We use the identity $\sin ^2 \frac{x}{2} = \frac{1}{2} (1-\cos x)$ and $\tan ^2 \frac{x}{2} = \frac{1-\cos x}{\sin x}$, then the above becomes
$\frac{1-\cos x}{2} \leq \frac{x^2}{4} \leq \frac{(1-\cos x)^2}{\sin ^2 x}$
$2 \leq \frac{x^2}{1-\cos x} \leq \frac{4(1-\cos x)}{\sin ^2 x} = \frac{4}{1+\cos x}$.
Squeezing gives the same answer.
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The next step is then to find the third order estimates. Or, when instincts kicks in, you would hope that this is what you need to generalize the whole thing. However, the deeper you look into the problem the bigger trouble you would find.
The use of quadratic related identities would fail instantly because you know it only give factors of (powers of) 2, which is not good if we are at higher orders. They are simply not good enough to prove the estimate of the next order. (But how did we managed to prove the second order estimate of $\cos x$? My rough guess is that the second order estimate of $\cos x$ is equivalent in some sense to the first order estimate of $\sin x$ although I am not so sure.)
For the other approach, you would wish that you can start with the formula $\sin ^n \frac{x}{n} \leq n^{-n}x^n \leq \tan ^n \frac{x}{n}$ and apply the multiple angle formula. I can see that being a possibility, albeit a very slim one.
The multiple angle formula can be written as $\sin x = T_n(\sin \frac{x}{n})$ where $T_n$ is the Chebyshev polynomial of the n-th order. The surprising thing is, equations in form of $T_n(x) + q = 0$ is solvable in radicals (!!!). That says, you can explicitly express $\sin \frac{x}{n}$ in a nested radicals in $\sin x$ solely and the estimation may proceed. Are you surprised that the name of Galois appears consecutively in my math entries by the way?
For $n = 3$, the triple angle formula induced equation $4x^3 - 3x + q = 0$ has the real solution $x = \frac{1}{2}(r + r^{-1})$ where $r = \sqrt[3]{\sqrt{q^2-1}-q}$. For our purpose we know $q = \sin x$ so that it even simplifies to $r = \sqrt[3]{i\cos x - \sin x}$. The root is real since $q \leq 1$, and we can simplify that to a single real expression in terms of $\sin x$ (left as exercise). BUT, how do we actually retrieve the term $x-\sin x$ from there? This is another big problem...
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That does not mean we are hopeless against such limit though.
The term $x- \sin x$ is still geometrically natural on the circle chart as the difference between length of arc $CB$ and the length of segment $CD$. Set midpoint of segment and arc $BC$ as $F$ and $G$ respectively. The arc length can then be bounded using the length $CB$ and $FG$ (note that $CB$ alone is not enough for the third order estimate!). Such approach works for third order estimates, but not any further when we can't find corresponding interpretation for higher order estimates on the chart.
Instead of trying hard with the trig circle, we just reside to the use the limit toolbox...as long as we know the limit exists, but that's easy right?
The existence can be done by continuity and MCT as long as it is bounded. First order bound gives $x-\sin x \leq \tan x - \sin x = \sin x (\sec x -1)$. Notice that $\lim _{x\to 0}\frac{\sec x - 1}{x^2} = \frac{1}{2}$ (why?), we conclude that $x-\sin x \leq (\frac{1}{2}+\varepsilon) x^3$ in a neighbourhood of $x = 0$.
With the existence of limit being shown, we have all the tricks in our sleeves. Here are two neat solutions I like. Set $L = \lim _{x\to 0} \frac{x-\sin x}{x^3}$.
Solution 1.
By triple angle formula:
$x-\sin x = x - 3\sin \frac{x}{3} + 4\sin ^3 \frac{x}{3}$
$= \frac{1}{9}(\frac{(x/3) - \sin (x/3)}{(x/3)^3}) + \frac{4}{27} \left ( \frac{\sin (x/3)}{x/3} \right )^3$
$\to \frac{1}{9}L + \frac{4}{27}$
which gives us $L = \frac{1}{6}$.
Solution 2.
Note that $L$ is also the limit of $\frac{2x - \sin 2x}{8x^3}$, a linear combination of limits shows that
$4L - L = 3L = \lim_{x\to 0}\frac{x - (1/2)\sin 2x - x + \sin x}{x^3} = \lim_{x\to 0}\frac{2\sin x - \sin 2x}{2x^3}$.
Now $\frac{2\sin x - \sin 2x}{2x^3} = \frac{\sin x}{x} \cdot \frac{1-\cos x}{x^2} \to \frac{1}{2}$, hence $L = \frac{1}{6}$.
It turns out that squeezing is trying to prove existence and value in one go which makes things strictly harder, the need of monotonic estimates is really hard to deal with as well. In the above approach, we only need $x - \sin x = \Theta (x^3)$, in contrast to the bound of $\frac{1}{6}x^3 + o(x^4)$ for squeezing. It is also note worthy that n-th order estimate always gives a (n+2)-th order bound (i.e. generalization is possible) because if the n-th coefficient matches, we can argue the (n+1)-th order coefficient is zero by odd parity.
We may need to admit that squeezing theorem has its limit(!) after all.
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Here is a bonus bound as suggested by Grok4 when I tested its capabilities, and it claimed that this is an 'elementary approach'. I asked it to prove $x - \sin x \leq \frac{1}{2}x^3$ and it returns as below:
Note Euler's infinite product $\sin x = x \prod (1 - \frac{x^2}{n^2 \pi ^2})$. Manipulating the formula gives
$\sin x = x \cdot \prod (1 - \frac{x^2}{n^2 \pi ^2}) \geq x(1 - \sum \frac{x^2}{n^2\pi ^2}) = x - \zeta(2) \frac{x^3}{\pi ^2}$,
and the claim follows since $\zeta (2) = \frac{\pi ^2}{6}$.
Beautiful? Yes. Elementary? Ummm...what's the difference between using infinite product and using Taylor series?
That isn't even the funniest part. In reality it failed to retrieve the tight constant $\frac{1}{6}$. Instead it uses the estimation $\sum \frac{1}{n^2} \leq 1 + \sum \frac{1}{n(n+1)} \leq 2$, so $\sin x \geq x - \frac{2}{\pi ^2}\cdot x^3$, where $\frac{2}{\pi ^2}\approx 0.2026$.
I asked is it possible to prove the inequality with the tight constant $\frac{1}{6}$ by elementary means then? It searched and thought seriously for a while, then it said no.
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