Can we prove $\lim _{x\to0}\frac{\sin x}x=1$ with a functional definition for $\sin(x)$?

Question

We all know the geometric proofs for $$\lim _{x\to0}\frac{\sin(x)}x=1$$

can we find one based on purely functional definition? For example, let's take Apostol's definition:

1. $\sin (x)$ and $\cos (x)$ are defined for all $x \in \mathbb R$.

2. $\cos (0) = \sin (\frac{\pi}2) = 1$ and $\cos (\pi) = -1$.

3. $\cos (y-x) = \cos y \cos x + \sin x \sin y$ for all $(x,y) \in \mathbb R ^2$.

and there is a fourth one used to get the limit:

4. $0 < \cos (x) < \frac {\sin (x)}x <1$ for $0 < x< \frac{\pi}2$.

Do we always need this fourth one? Is that limit really unobtainable without geometry?

It seems to me we can't reach all values of sine and cosine with just the three properties, as we can only define $\aleph _0$ values for $\cos (x)$, but could we change the fourth one for "$\cos(x)$ is continuous" or something similar and derive the fundamental limit?

Notice that in the use of the Taylor expansion we're already assuming the limit to be one as we are using the derivative of $\sin(x)$.

Answer 1

This question is related to this one. Any definition that allows a uniqueness proof should let us prove the desired limit. It turns out that continuity is not required for this. So long as we carefully specify the wavelength in terms of $\pi$, along with the Pythagorean and angle sum and identities, we can prove uniqueness. In particular, it can be proven that there exists a unique pair of functions $(\cos,\sin)$ mapping $\mathbb{R}\to\mathbb{R}$ which satisfy the following conditions.

1. The Pythagorean identity $\cos(x)^2+\sin(x)^2=1$
2. The angle sum formulae $$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$ $$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$$
3. The wavelength condition, $\sin(\pi)=0$ while $0<\sin(x)$ for any $x\in(0,\pi)$

By proving that this characterization is unique, and that the ordinary Taylor series for $(\cos,\sin)$ satisfies all these conditions, we prove the desired limit and much more. The proofs of these claims can be found below.

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We first show that the functions $(\cos,\sin)$, if they exist, are unique. To do this, we translate our four conditions into a more manageable state. Define the function $H(x)=\cos(x)+i\sin(x)$, which has real part $\Re(H(x))=\cos(x)$ and imaginary part $\Im(H(x))=\sin(x)$. Using this definition, our conditions are point by point equivalent to the following.

1. $|H(x)|=1$
2. $H(x+y)=H(x)H(y)$
3. $H(\pi)\in\mathbb{R}$ while $\Im(H(x))>0$ for any $x\in(0,\pi)$.

To show $(\cos,\sin)$ is unique, we instead prove that $H$ is unique. Since $H(\pi)\in\mathbb{R}$ has $|H(\pi)|=1$, then $H(\frac{\pi}{2})^4=H(\pi)^2=1$, so that $H(\frac{\pi}{2})\in\{1,i,-1,-i\}$. Since $\Im(H(\frac{\pi}{2}))>0$, then $H(\frac{\pi}{2})=i$. Using this, we find that $H(x+\frac{\pi}{2})=H(x)H(\frac{\pi}{2})=iH(x)$, thus $\sin(x+\frac{\pi}{2})=\cos(x)$ and likewise $\cos(x+\frac{\pi}{2})=-\sin(x)$. It follows that for $x\in(0,\frac{\pi}{2})$ we have $\sin(x)>0$ and $\cos(x)=\sin(x+\frac{\pi}{2})>0$, thus $H(x)$ is in the first quadrant. For $x\in(\frac{\pi}{2},\pi)$, we have $H(x-\frac{\pi}{2})=-iH(x)$ in the first quadrant, thus $H(x)$ is in the second quadrant. Continuing like this, any $x\in(\pi,\frac{3\pi}{2})$ has $H(x)$ in the third quadrant, and likewise any $x\in(\frac{3\pi}{2},2\pi)$ has $H(x)$ in the fourth quadrant. Since $H(2\pi)=H(\pi)^2=1$, then $H(x+2\pi)=H(x)$, so the quadrant of $H(x)$ is entirely determined by $x$.

Let $H_1,H_2$ be two functions satisfying our conditions, then we prove $H_1=H_2$. Let $I(x)=\frac{H_1(x)}{H_2(x)}$, and notice that $|I(x)|=\frac{|H_1(x)|}{|H_2(x)|}=1$, and likewise $I(x+y)=\frac{H_1(x)H_1(y)}{H_2(x)H_2(y)}=I(x)I(y)$. Therefore if we let $C(x)=\Re(I(x))$ and $S(x)=\Im(I(x))$, the pair $(C,S)$ satisfy our conditions 1 and 2. Since $H_1$ and $H_2$ are always in the same quadrant, and $H_1=H_2\cdot I$, we must have $\Re(I(x))>0$ for all $x$, thus $C(x)>0$. Let $\sigma$ be the infimum value of $C(x)$, and notice the following. $$C(x) \geq C(x)^2 = \frac{(2C(x)^2-1)+1}{2} = \frac{C(2x)+1}{2} \geq \frac{\sigma+1}{2}$$

The first inequality holds since $C(x)\in (0,1]$, where $C(x)\leq 1$ follows from the Pythagorean identity. The equalities after that follow from a basic rearrangement of the double angle formula (derived from conditions 1 and 2 on $I$). The last inequality follows from $C(2x)\geq \sigma$. Since $\sigma$ is the infimum of $C(x)$, and $C(x)\geq\frac{\sigma+1}{2}$, it follows that $\sigma\geq \frac{\sigma+1}{2}$, which is equivalent to $\sigma\geq 1$. It follows that $C(x)\geq 1$ for all $x$, but since $C(x)\leq 1$ then $C(x)=1$ for all $x$, therefore $S(x)^2=1-C(x)^2 = 0$ so that $I(x)=1$ is constant, proving $H_1=H_2$.

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To finish our demonstration, let $\cos,\sin$ be defined by the usual Taylor series, then we show that these satisfy our conditions, and are thus the unique solution. As some preliminary observations, notice that by the Taylor series we have $\sin(-x)=-\sin(x)$ and $\cos(-x)=\cos(x)$. Let $\exp$ be the natural exponential function defined by its power series for any complex number, which has $\exp(0)=1$. Euler's formula comes out very quickly from this, that $\exp(ix)=\cos(x)+i\sin(x)$. We can also prove the product formula for exponentials $\exp(x+y)=\exp(x)\exp(y)$, which works even if $x,y$ are complex numbers.

A simple consequence of the above observations is that $\exp(i(x+y))=\exp(ix)\exp(iy)$. Since $H(x)=\cos(x)+i\sin(x)=\exp(ix)$, this proves $H(x+y)=H(x)H(y)$, condition 2, the angle sum formulas. Applying the negative angle formulas, we find that $H(-x)=\overline{H(x)}$ is the complex conjugate, thus $1=H(0)=H(x)H(-x)=|H(x)|^2$ proves condition 1, the Pythagorean identity.

Previously we showed that if some function $I$ satisfied our conditions 1 and 2, and also obeyed $\Re(I(x))>0$, then $I$ would be constant. By the contrapositive, since $H$ is not constant, $H$ fails one of the three premises. Since $H$ satisfies conditions 1 and 2, it must be the case that some $x$ has $\Re(H(x))\leq 0$, that is to say, $\cos(x)\leq 0$. Since $\cos$ is even, we can assume that $x\geq 0$. Now we can take the infimum $\ell=\inf\{x\geq 0 : \cos(x)\leq 0\}$. Due to continuity and $\cos(0)=1$, we must have $\ell>0$. Taking the limit from the right side, $\cos(\ell)\leq 0$, but taking a limit from the left side, we instead get $\cos(\ell)\geq 0$, so in fact $\cos(\ell)=0$. Now we get the following. $$\sin(2\ell)=2\sin(\ell)\cos(\ell)=0$$

As above, there exists $x>0$ for which $\sin(x)=0$, namely $x=2\ell$. Let $\lambda=\inf\{x>0 : \sin(x)=0\}$, and notice that since $\lim_{x\to 0}\frac{\sin(x)}{x}=1>0$ by the Taylor series, then $\lambda>0$. By continuity we get $\sin(\lambda)=0$, and since $\sin$ is positive in a right-side neighborhood of $0$, then all $x\in(0,\lambda)$ must have $\sin(x)>0$. Now we just prove $\lambda=\pi$.

Recall that $H(x)=\cos(x)+i\sin(x)$ has $|H(x)|=1$ for all $x$, thus $H$ stays on the unit circle. More specifically for $x\in(0,\lambda)$ we have $\Im(H(x))>0$, so on this range, $H$ stays on the upper half unit circle. Analyzing the taylor series, we find that $\frac{d}{dt}\cos(t)=-\sin(t)$, so for $t\in(0,\lambda)$ we have $\frac{d}{dt}\cos(t)=-\sin(t)<0$, thus $\cos$ is strictly decreasing, so $H(x)$ is injective over $t\in[0,\lambda]$. Since $H(0)=1$ while $H(\lambda)=-1$, by continuity $H(x)$ is surjective to the upper half unit circle over $x\in[0,\lambda]$. It follows that over the range $[0,\lambda]$, we have $H$ smoothly and bijectively parametrizing the upper half unit circle. We can therefore calculate $\pi$ as follows. $$\pi = \int_0^\lambda \left|\frac{d}{dt}H(t)\right|dt = \int_0^\lambda \left|\frac{d}{dt}\exp(it)\right|dt = \int_0^\lambda |i\exp(it)|dt = \int_0^\lambda 1dt = \lambda$$

Indeed, if we consider $t$ to be time, then $H(t)$ traces the upper half unit circle, with its velocity being the derivative $\frac{d}{dt}H(t)$. The magnitude of the velocity is its speed, so by integrating over the speed, we produce the total length of the path traveled, which is the arclength of the upper half unit circle, exactly $\pi$. Since the integral reduces to $\lambda$, then $\lambda=\pi$, proving that our $(\cos,\sin)$ satisfy condition 3.

Answer 2

No, you do not need to assume the specific limit. However, the simple functional equations for the sine and cosine functions, alone, are not enough to make for a unique solution, even together with continuity.

The reason for this is that the conditions, or any equivalent of

1. Initial conditions: $\cos(0) = 1$ and $\sin(0) = 0$
2. Pythagorean identity: $[\cos(x)]^2 + [\sin(x)]^2 = 1$
3. The angle-sum formulas: $$\cos(x + y) = \cos(x) \cos(y) - \sin(x) \sin(y)$$ $$\sin(x + y) = \cos(x) \sin(y) + \sin(x) \cos(y)$$
4. Continuity.

are enough to specify the shape of the functions, i.e. that they are sine waves, and they are also enough to specify the amplitude. What you cannot derive from this is the wave frequency.

Here's how that works. The most natural way to motivate taking (2) and (3) as axiomatic - and why I prefer this axiom set over the one you mention - is that they are best understood via the concept of rotations, which can be represented as matrices. In particular, $\cos$ and $\sin$ are the functions that parameterize a rotation matrix in terms of the angle of rotation:

$$R(\theta) := \begin{bmatrix}\cos(\theta) && -\sin(\theta) \\ \sin(\theta) && \cos(\theta)\end{bmatrix} \cdot$$

where the $\cdot$ at the end is not a period, but meant to signify matrix multiplication, i.e. $R(\theta)$ is an operator acting on 2-vectors. Rotations, themselves, actually do not pre-require the trig functions: a rotation as a geometric transformation after all is defined by its fact of preserving the Euclidean distance as well as chirality (the distinction between left and right, which excludes mirror transformations). Instead, seen this way, the $\cos$ and $\sin$ functions serve as a certain way of organizing those rotations - and thus one should expect there may be alternative ways. In particular, the two formulas in Axiom (3) are equivalent to postulating that

$$R(\theta + \phi) = R(\phi) R(\theta)$$

i.e. that summation of the angles should be the same as composition of the rotations. Likewise, we still need to obtain axiom (2), but it is exactly the statement that the determinant of this matrix is $1$, which means the rotation contains no nontrivial rescaling component.

However, now you may see the problem: we have not specified a unit with which to measure the angles! In fact, rotations and angle measures always compose like this. If the arguments to $\cos$ and $\sin$ were to be interpreted as angles in degrees, you would still find that the angle sum formulas hold, as do the initial conditions and continuity! Or consider a rotating wheel - you wait some time then wait some more, that's a composition of rotations, and you expect that it doesn't matter that you punctuate your waiting, the rotations should still add to give the total rotation as if there were no punctuation at all. The trick now is to see this doesn't depend on the rotation speed.

Formally, it is also quite easy to see - just consider $\cos^{*}(x) := \cos(fx)$ and $\sin^{*}(x) := \sin(fx)$. Then you can do

$$\begin{align} \cos^{*}(x + y) &= \cos(f(x + y))\\ &= \cos(fx + fy)\\ &= \cos(fx) \cos(fy) - \sin(fx) \sin(fy)\\ &= \cos^{*}(x) \cos^{*}(y) - \sin^{*}(x) \sin^{*}(y) \end{align}$$

and in like manner for $\sin^{*}$. Hence solution of the angle formulas, itself, does not determine the scale factor $f$, because any such scaling also satisfies them!

So how can we do it? Well, we must add at least one more axiom. Interestingly, that doesn't have to be a limit. I believe that adducing the axiom

5. Define the constant $\tau$ as the arc length of a unit circle. $$\tau := 4\int_{0}^{1} \frac{dx}{\sqrt{1 - x^2}}$$ Then require that $\tau$ is both the least period of $\sin$, and that $\sin\left(\frac{\tau}{4}\right) = 1$.

is sufficient. What this is doing is in effect setting our angular measure to be similar to linear measure, i.e. it is measurement of actual length, on the circle with radius of one unit. This, of course, is the definition of radian measure. The "least period" part is vital because, as you point out, merely specifying this point alone still wouldn't be enough as we would have to contend with the existence of an infinite number of equivalent harmonics. Finally, the additional stipulation on the value at $\frac{\tau}{4}$ is necessary to set the sense of the rotation to counterclockwise.

Hence, you cannot prove uniqueness of $\cos$ and $\sin$ without another axiom of this type. But that, of course, is a bit stricter than finding a particular limit. However, the trick with the limit

$$\lim_{x \rightarrow 0} \frac{\sin x}{x}$$

is that it is actually one and the same as the derivative of $\sin$ at $0$, i.e. $\sin'(0)$. And it turns out that is indeed equivalent to measuring the frequency.

The whole problem is analogous, pretty much, to that of defining exponential functions by functional equations - in particular, exponential functions depend on a base, and likewise here the "base" is frequency. In fact, the two are essentially wholly equivalent when you pass them through the complex number domain, because there exponentials and trigonometric functions are simply "imaginary" versions of each other (namely $\cos$ and $\sin$ are the imaginary parts of $x \mapsto e^{ix}$, via Euler's formula).

Answer 3

Let's first work a bit without $(4)$:

$(3)$ with $y=x$ together with $(2)$ gives us $$ \tag a \cos^2x+\sin^2x=1.$$ In particular, $\sin 0=\sin\pi=0$ and $\cos\frac\pi2=0$. Next, $(3)$ with $y=0$ gives $$\tag b \cos(-x)=\cos x.$$ With $y=\frac \pi2$, $$\tag c\cos (\frac\pi2-x) = \cos \frac\pi2\cos x+\sin \frac\pi2\sin x=\sin x $$ and with $y=\pi$, $x=\frac\pi2-t$, $$\tag d \cos(\frac\pi2+t)=\cos\pi\cos(\frac\pi2-t)+\sin \pi\sin(\frac\pi2-t)=-\cos(\frac\pi2-t)=-\sin t. $$ Next, with $y=\pi$, $x=-t$ $$\tag e \cos(\pi+t)=\cos\pi\cos(-t)+\sin\pi\sin(-t)=-\cos(-t)=-\cos t$$ and by repeating, $$ \tag f\cos(2\pi+x)=\cos x.$$ By (d), this also implies $$\tag g \sin (\pi+x)=-\sin x,\qquad \sin (2\pi+x)=\sin x. $$ From (c) and (d), we also have $$\tag h \sin (-x)=-\sin x. $$

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Now we can use $(4)$ in the following weakened form:

$$\tag{4'}\exists \epsilon>0\colon \forall x\in (0,\epsilon)\colon\; 0<\cos x<\frac{\sin x}x<1.$$

From $(4')$, we also see $0<\sin x<x$ for sufficiently small positive $x$, hence $$\lim_{x\to 0^+}\sin x=0,$$ so from (a), $$\lim_{x\to 0^+}\cos^2 x=1.$$ As $\cos x>0$ for sufficiently small positive $x$, this implies $$\lim_{x\to 0^+}\cos x=1.$$ Then by squeezing, $(4')$ gives us $$\lim_{x\to0^+}\frac{\sin x}x=1. $$ From this and (h), $$\lim_{x\to0}\frac{\sin x}x=1. $$

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On the other hand, if we drop $(4)$ completely, we know that for example $\sin 5x$ and $\cos 5x$ in place of $\sin x$, $\cos x$ solfe the functional equation while giving another limit. We conclude that it is not so much the sign of sine or cosine on all of $[0,\frac\pi2)$ that is important here, but (apart from the signs on an arbitrarily small interval) rather the fact that $\frac{\sin x}x$ does not exceed $1$.

At first glance, $(4)$ looks like a very innocent inequality, way weaker than, say, assuming continuity. Then again it was good enough to obtain a (one-sided) limit for sine at one point as a milestone. In fact, if we drop $(4)$, much weirder solutions exist beyond just higher frequencies: Let $f$ be a non-continuous solution of the Cauchy functional equation $$f(x+y)=f(x)+f(y)$$ such that $f|_{\pi\Bbb Q}$ is the identity (or a $1\bmod 4$ multiple thereof). Then $\sin\circ f$, $\cos\circ f $ solve $(2)$, $(2)$, $(3)$.