Is there anything wrong with this proof that $\lim_{x\to0} \frac{\sin(x)}{x} = 1$

Question

Definitions

Since I'm asking for verification of a proof, I'll start with some basic definitions. I believe these definitions are more or less equivalent to the "standard" definitions of curve length, radian, and sine/cosine:

• Let the length of a continuous curve $f(t): [a,b] \rightarrow \mathbb{R}^2$ be defined as $\lim_{N\to\infty}\sum_{i=1}^N \sqrt{(x(t_{i})-x(t_{i-1}))^2 + (y(t_{i})-y(t_{i-1}))^2}$ where $(x(t_{i}),y(t_{i}))=f(t_{i})$ and $t_{i} = a+i(b-a)/N$.
• Let the angle measured by the counter-clockwise rotation of a point $P$ on a unit circle $C$, such that the locus of $P$ generates a curve of length $\theta$, be defined as $\theta$ radians. This angle measured by a clockwise rotation can be defined as $-\theta$ radians.
• Let the point on the Cartesian plane reached by rotating the point $(1,0)$ on the unit circle centered at $(0,0)$ by an angle of $\theta$ radians be defined as $(\cos(\theta), \sin(\theta))$.

Detailed Proof

Consider the curve $f(t) = (\cos(t), \sin(t))$ for $t \in [0,1]$. On the one hand, this is just a circular arc that subtends 1 radian from the center of a unit circle, and must therefore have length 1, by the definition of a radian.

On the other hand, we can compute the length of the curve by the definition of curve length, which means that: $$\lim_{N\to\infty}\sum_{i=1}^N \sqrt{(\cos(i/N)-\cos((i-1)/N))^2 + (\sin(i/N)-\sin((i-1)/N))^2} = 1$$ Using basic trigonometric properties and algebra (see appendix for details), we can get: $$\lim_{N\to\infty}2N\sin(\frac{1}{2N}) = \lim_{N\to\infty}\frac{\sin(\frac{1}{2N})}{\frac{1}{2N}} = 1$$ Since we can always choose an $x\in\mathbb{R}$ such that $0<x<\frac{1}{2N}$, we have that: $$\lim_{x\to0^+}\frac{\sin(x)}{x} = 1$$ Now we just observe that $\sin(x)$ is an odd function since we rotate starting from the x-axis, and therefore a counter-rotation reflects the y-coordinate along the x-axis, so we have: $$\lim_{x\to0^-}\frac{\sin(x)}{x} = \lim_{x\to0^+}\frac{\sin(-x)}{-x} = \lim_{x\to0^+}\frac{-\sin(x)}{-x} = \lim_{x\to0^+}\frac{\sin(x)}{x} = 1$$ Therefore: $$\lim_{x\to0}\frac{\sin(x)}{x} = 1$$ QED

Note, however, that to show that the length of the curve is $\lim_{N\to\infty}2N\sin(\frac{1}{2N})$, we don't need to use trigonometric properties or algebra at all. Consider the following diagram where we break a circular arc of length 1 into N=3 parts:

The summation in the definition of the curve length is just the sum of the lengths of the orange segments in the above diagram. Notice that all the orange segments are of equal length since all the triangles are congruent isosceles triangles with two side lengths of 1 joined at an angle of 1/N = 1/3 radians. Therefore, the sum of the lengths of all segments is 2N multiplied by half the length of a single segment. To find half the length of a single orange segment, we split the bottom triangle into two right triangles and rotate the bottom half like so:

This shows that half the length of a segment is $\sin(1/6)$ for N=3 and $\sin(\frac{1}{2N})$ in general. Therefore, the general the sum of the lengths is $2N\sin(\frac{1}{2N})$.

TL;DR Proof

If I was showing this as a basic proof, with the same level of rigor as I usually see the proof of $\lim_{x\to0} \frac{\sin(x)}{x} = 1$ thrown around everywhere, it would be much more straightforward:

Here's a diagram splitting an arc of length 1 into N equal parts:

Observe that there are N parts each of length $2\sin(\frac{1}{2N})$, so the total length is $2N\sin(\frac{1}{2N})$. This implies that: $$\lim_{N\to\infty}2N\sin(\frac{1}{2N}) = \lim_{N\to\infty}\frac{\sin(\frac{1}{2N})}{\frac{1}{2N}} = 1$$ Now let $x = \frac{1}{2N}$ and we get: $$\lim_{x\to0}\frac{\sin(x)}{x} = 1$$ QED

Motivation

Why do I bring up this proof? If it's valid, I honestly think this is a much better proof than the common proof that relies on the squeeze theorem and geometric areas demonstrating that $\frac{\sin(\theta)}{2}\leq\frac{\theta}{2}\leq\frac{\tan(\theta)}{2}$. The common proof hides the fact that we ultimately need to find the limit of an infinite process. To do this, it relies on the area of a circular sector.

Ironically, many common proofs that we are taught for the area of a circle, if expressed more rigorously, are actually relying on this very same limit. You can even see this in the above diagrams. If we split a sector of arc length $\theta$ into N triangles, each one has area $\frac{1}{2}\sin(\frac{\theta}{N})$ and therefore the total area is: $$\lim_{N\to\infty}\frac{1}{2}N\sin(\frac{\theta}{N}) = \lim_{N\to\infty}\frac{1}{2}\frac{\sin(\frac{\theta}{N})}{\frac{1}{N}} = \frac{1}{2}\lim_{x\to0}\frac{\sin(\theta x)}{x}$$

And it is precisely the proof shown here that can be trivially extended to show that: $$\lim_{x\to0}\frac{\sin(\theta x)}{x} = \theta$$

So in short, I believe the advantages of using the proof here, if valid, as the typical proof for demonstrating that $\lim_{x\to0} \frac{\sin(x)}{x} = 1$ are:

1. It avoids circular reasoning if we're relying on a modern notion of Archimedes' proof for the area of a circle.
2. It avoids dealing with areas at all and instead directly deals with the ratio of two measures of length.
3. It easily extends to a more general result, namely that $\lim_{x\to0} \frac{\sin(\theta x)}{x} = \theta$.

Appendix

$$\lim_{N\to\infty}\sum_{i=1}^N \sqrt{(\cos(i/N)-\cos((i-1)/N))^2 + (\sin(i/N)-\sin((i-1)/N))^2}$$ The above expression will contain $(\cos(i/N)^2 + \sin(i/N)^2)$ and $(\cos((i-1)/N)^2 + \sin((i-1)/N)^2)$ so once we simplify we will get the expression: $$\lim_{N\to\infty}\sum_{i=1}^N \sqrt{2 – 2(\cos(i/N)\cos((i-1)/N) + \sin(i/N)\sin((i-1)/N))}$$ Since cosine is an even function and sine is an odd function, this is equivalent to: $$\lim_{N\to\infty}\sum_{i=1}^N \sqrt{2 – 2(\cos(i/N)\cos((1-i)/N) – \sin(i/N)\sin((1-i)/N))}$$ Since $\cos(a+b) = \cos(a)\cos(b) – \sin(a)\sin(b)$, we get the expression: $$\lim_{N\to\infty}\sum_{i=1}^N \sqrt{2 – 2(\cos(i/N + (1-i)/N))} = \lim_{N\to\infty}\sum_{i=1}^N \sqrt{2 – 2\cos(1/N)}$$ Since $1 – \cos(\theta) = 2\sin^2(\theta / 2)$, we get: $$\lim_{N\to\infty} N\sqrt{2*2\sin^2(\frac{1}{2N})}$$ And the final result: $$\lim_{N\to\infty} 2N\sin(\frac{1}{2N})$$

Answer 1

A crucial step in your reasoning is:

[...] since we can always choose an $x\in\mathbb{R}$ such that $0<x<\frac{1}{2N}$, we have that: $$\lim_{x\to0^+}\frac{\sin(x)}{x} = 1$$

I am not sure about this step; the implication does not follow. You have shown that as one sequence (namely, $1/2N$) tends to $0^+$, the function tends to $1$. It is necessary to show the limit is $1$ for every sequence tending to $0^+$.

To complete your proof, it would suffice to demonstrate the (true) result that:

$$x \mapsto \frac{\sin(x)}{x}$$

is monotonically increasing as $x \to 0^{+}$.