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$$\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1$$ 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?

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    $\begingroup$ is your teacher serious about this statement? $\endgroup$ – tired Nov 9 '15 at 11:23
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    $\begingroup$ what is the definition of $\sin ?$ $\endgroup$ – abel Nov 9 '15 at 11:27
  • $\begingroup$ You should ask your teacher why she keeps teaching junk then... $\endgroup$ – mattecapu Nov 9 '15 at 12:02
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    $\begingroup$ If one draws a circle in the course of the proof the proof is certainly circular. $\endgroup$ – Christian Blatter Nov 9 '15 at 12:09
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    $\begingroup$ It is true that some proofs of this result given in elementary calculus texts are circular (for example, anything which uses the fact that $\sin$ is differentiable at zero with derivative 1, so any proof using the McLaurin series, has already used the existence of that limit) but I would not normally expect a fully rigorous proof to be given at high school level. Some proofs which purport to be rigorous rely on pictures of circles, so you have to accept some "visually obvious" facts about circles. $\endgroup$ – Geoff Robinson Nov 9 '15 at 12:22
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$$\sin t=\frac{e^{it}-e^{-it}}{2i}.$$

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.

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  • $\begingroup$ If $\sin x$ is defined through an infinite series could I just ignore the higher-order terms and get $\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1$? After which is the burden of proof lies in relating the function to its geometrical meaning? $\endgroup$ – MarcoXerox Nov 9 '15 at 13:53
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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 $C=\frac\pi{360°}$.

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  • $\begingroup$ I see your point and in fact it is close to the proof in my textbook. However, I'd like to know if the equation $s = r\theta$ is proven through the limit $\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1$ I've tried to work it out both ways but I'd really love to know which comes from which. $\endgroup$ – MarcoXerox Nov 9 '15 at 13:51
  • $\begingroup$ It is more advanced and complicated to argue by arc length and polygonal approximations of the arc. Comparing areas just invokes subset relations. $\endgroup$ – Lutz Lehmann Nov 9 '15 at 14:34
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    $\begingroup$ While this isn't circular per se, it does involve circles. $\endgroup$ – Akiva Weinberger Nov 9 '15 at 18:16
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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?

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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.

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  • $\begingroup$ If you "prove" that formula for $\sin$, you most likely have already proved before that $\sin \theta /\theta \rightarrow 1$. $\endgroup$ – Aloizio Macedo Nov 9 '15 at 11:46
  • $\begingroup$ You using "prove" indicates that you may not believe the "formula" is true. @AloizioMacedo $\endgroup$ – Megadeth Nov 9 '15 at 11:57
  • $\begingroup$ No, it indicates that you used the term. And you didn't address the issue: how do you prove that formula without already knowing that $\sin \theta /\theta \rightarrow 1$? My point is: either that formula is your definition, or if you prove it you most likely have already proved before that $\sin \theta/ \theta \rightarrow 1$. $\endgroup$ – Aloizio Macedo Nov 9 '15 at 12:06
  • $\begingroup$ @AloizioMacedo Ah, proving that $\sin$ is analytic need not employ $(\sin \theta)/\theta \to 1$. $\endgroup$ – Megadeth Nov 9 '15 at 12:11
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    $\begingroup$ The fact that $\lim_{ x \to 0} \frac{\sin{x}}{x}= 1$ is equivalent to the fact that $\sin^{\prime}(0) = 1$, so you do need that limit to obtain the McLaurin series for $\sin$. $\endgroup$ – Geoff Robinson Nov 9 '15 at 12:26
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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?

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  • $\begingroup$ I currently know that the function $f(x) = \sum_{n = 0}^{\infty} (-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$ comes from Maclaurin series of $\sin x$. Is it true that $\sin x$ is actually defined as an infinite series and being related to an infinite series in modern maths, which is not covered by high school syllabus? $\endgroup$ – MarcoXerox Nov 9 '15 at 13:32
  • $\begingroup$ That is one way to define it, but there are others. $\endgroup$ – Geoff Robinson Nov 9 '15 at 13:34

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