Non-circular proof of $\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1$

Question

$$\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1$$ The above limit is fundamental to studies of introductory calculus. I know that this limit could be proven by the squeeze theorem and the length of sector, i.e. $$s = r\theta$$ where r is radius and $\theta$ the angle. However, it is claimed that the proof of this limit is circular.

I can't bring myself to agree to that, but apparently the length of sector is a corollary of the limit, which is proven by the inequality $$\cos \theta < \frac{\sin\theta}{\theta} < 1$$ Can anyone point me to other proofs of the limit?

Answer 1

$$\sin t=\frac{e^{it}-e^{-it}}{2i}.$$

Hence,

$$ \frac{\sin t}{t}= \frac{1+it+\epsilon_1(t)-1+it+\epsilon_2(t)}{2it}=\frac{2it+\epsilon(t)}{2it}=1+\frac{\epsilon(t)}{2it}, $$

where $\epsilon(t)/2it \rightarrow 0$ as $t \rightarrow 0$.

There is no circularity: We don't "use the derivatives of $\sin$ to arrive at the Taylor series". It is the definition (one of the possible ones, but this is one good to work with). Motivating the definition is another issue altogether.

For instance, you could prove that there exists at most one pair of functions $s,c$ such that $s'=c$, $c'=-s$, $s^2+c^2\equiv 1$ and $s(0)=0$. These properties are easy to "prove" for the case where $\sin$ is "geometrically" defined (the definition of this is by drawing... so the best you can get is arguments by drawing too. It is not circular, it is just ill-founded) and easy to prove from the definition above. Hence, they correspond.

Answer 2

Usually the easiest to understand geometric justification is via the area of the sector of the unit disk associated with $θ$. What should be clear by the geometric definition of the angle is that this area is proportional to $θ$, $A(θ)=Cθ$.

Using triangles inside the sector you get a lower bound of either $\frac12\cosθ\sinθ=\frac14\sin2θ$ or $\sin\fracθ2\cos\fracθ2=\frac12\sinθ$. Using triangles outside the unit disk, one gets upper bounds of either $\frac12\tan θ$ or $\tan \fracθ2=\frac{\sinθ}{1+\cosθ}$. Thus whatever the measure for the angle is, you get $$ \frac12\sinθ\le Cθ\le \frac{\sinθ}{1+\cosθ}\iff C·(1+\cosθ)\le\frac{\sinθ}{θ}\le 2C $$ which results in $\lim_{θ\to0}\frac{\sinθ}{θ}=2C$.

What value to assign to $C$ and closely related what measurement to assign to the angle of the full circle has now entirely analytical reasons to keep certain series from being cluttered with arbitrary constants. Without that it is completely reasonable to assign $θ=360°$ to the full circle and have $C=\frac\pi{360°}$.

Answer 3

Of course, the problem is the definition of the sinus. A possible answer is this.

Assume you are running on the unit circle at the constant speed one, starting from $(1,0)$. Let $(x(t), y(t))$ your position at instant $t$. Then $x(t)x'(t)+y(t) y'(t)=0$ (differentiate $x^2+y^2=1$), with $x'^2(t)+y'^2(t)=1$ (speed 1). It follows that $x'(0)= 0$, $y'(0)=\pm 1$. The sign $1$ means that you start in inverse sense of a watch. Thus the limit $\lim _{t\to 0} {y(t)\over t}= 1$. In fact $y(t$) is not a stupid definition for $\sin t$.

The proof is circular perhaps not in the sense you meant?

Answer 4

Wrong is using "the" in the sentence "... because the proof of this limit is circular..."; usually a statement admits more than just one proof. Besides, if a claimed proof is circular, then that simply means that it is false; for the person who wrote the proof was simply not playing the "game".

If you read Apostol's calculus, you can find a proof of the statement there. He uses four statements stipulating the behavior of $\cos$ and $\sin$ as axioms, one of which is $$ 0 < \cos \theta < \frac{\sin \theta}{\theta} < \frac{1}{\cos \theta}, $$ valid for all $\theta \in ]0,\pi/2[$; then it is easily concluded (check the book for this claim and the relevant details) that $$ \lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1. $$

Another way that came to my mind now is: It can proved that for all $n \geq 0$ we have $$ \sin \theta = \sum_{k=0}^{n}\frac{(-1)^{k}\theta^{2k+1}}{(2k+1)!} + o(\theta^{2n+1}) $$ as $\theta \to 0$; in particular, we have $$ \sin \theta = \theta + o(\theta) $$ as $\theta \to 0$; hence $$ \frac{\sin \theta}{\theta} = 1 + o(1) \to 1 $$ as $\theta \to 0$.

I have shown you a circular proof.

Answer 5

This is quite a difficult problem to address fully rigorously. It is possible ( with a lot more theory behind you) to define a function $f(x) = \sum_{n = 0}^{\infty} (-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$, and show that $f(x)$ exists for all real $x$ because the series converges everywhere. After the fact, we know that $f(x) = \sin{x}$, but if we use this as the definition of a function called $\sin$, how do we relate it to trigonometric functions, and how do we see (directly) that it is periodic?