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The same question have been asked here. But almost all the answers given there use the idea of squeezing one way or another, even this geometric proof uses the idea of squeezing. So, here is my question; how to prove
$$ \lim_{x\to 0}\frac{\sin x}{x}=1 $$ without the idea of squeezing involved.

Edited: No offense, but proofs of limits using integrals or derivative(or any other concepts that are defined using limits) to me is like building a house starting from the roof. So I am not looking for those kind of proofs.

marya
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  • Use the expansion of Taylor series for $\sin$. – Jean Marie Nov 24 '17 at 06:59
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    @Jean Marie For which we need firstly to calculate this limit. – Michael Rozenberg Nov 24 '17 at 07:04
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    Taylor series, L'hospital's rule, are usual approaches. On the other hand the answer depends on your definition of $\sin$ function. – Bumblebee Nov 24 '17 at 07:07
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    The initial assumption of this post is false: in the linked question there is plenty of answers that do not use "geometric" squeezing. Of course, some of them do not use geometric definitions. However, I'd like to point out that, even at "algebraic" level, "squeezing" (or, with a more appropriate term, "making estimates") is a perfectly fine and often essential tool of the evaluation of limits. Therefore, it is apparent that a geometric question will need some kind of geometric estimate. –  Nov 24 '17 at 07:10
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    Your note is rather funny because, formally, limits are far easier to define than a simultaneous notion of length for both line segments and circular arcs (which in fact requires a limiting process). Therefore, as difficulty goes: $$\text{limits}\ll \text{arc-length}\le \text{radiant}\le \text{trig functions}$$ –  Nov 24 '17 at 07:22
  • @G.Sassatelli: So, are you saying the proof I'm looking for is impossible? – marya Nov 24 '17 at 07:29
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    What definition of $\sin x$ are you using? -- In the end, proving $\frac{\sin x}x\to 1$ by definition means that you have to show that the fraction is "squeezed" between $1-\epsilon$ and 1$ – Hagen von Eitzen Nov 24 '17 at 07:45
  • @HagenvonEitzen: the usual, using sides of triangles. – marya Nov 24 '17 at 07:49
  • many the possible way to answer that question are already include in that post – Guy Fsone Nov 24 '17 at 15:01
  • The limit in question is an immediate consequence of the definition of symbol $\sin x$ and you should probably also mention your preferred definition. More likely when you will try to provide a definition you will get the answer to your question instantly. – Paramanand Singh Nov 24 '17 at 20:52
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    Btw using sides of triangle does not give a complete definition of symbol $\sin x$. For example if $x=1/2$ how do you use your triangle stuff to get $\sin (1/2)$? You will find that the approach using triangles has many hidden aspects which are difficult to handle. – Paramanand Singh Nov 24 '17 at 20:56

3 Answers3

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Use Taylor expansion of $\sin$ around zero. You have

$$\sin(x) = x+o(x)\,,$$

from which that limit follows immediately.

Paolo Intuito
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    Unfortunately, for a Taylor expansion, you need derivatives at $x=0$, and the first derivative at $x=0$ is exactly that limit. Of course, that depends on your definition of trigonometric functions. With the usual, geometric definition, you have a logical circle. If you define them by their power series, or by differential equations, there is no circle, and the problem is trivial. But you'd have to show that the so defined functions indeed have their role in geometry, then (not difficult, but not trivial, either). –  Nov 24 '17 at 07:13
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    First derivative of sine is $\cos(x)$.. Anyhow, it's plenty of Calculus 1 courses where sine and cosine are defined as power series. – Paolo Intuito Nov 24 '17 at 07:18
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    @G.S.: First derivative of $\sin$ is $\cos$ provided that you be able to compute it using limits - which is exactly what the question is trying to achieve. You are going in circles about it. Don't feel encouraged by the 2 upvotes received, they have been cast by people with little understanding of the matter and they could be easily reverted by an even larger number of downvotes. – Alex M. Nov 24 '17 at 14:25
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You can find the limit using geometry! It emerge when calculating the limit of the area of a regular $n$-gone inscribed on a circle of radius $r$. The limit of the area as $n\to\infty$ must be equal to the area of the circle $\pi r^2$. Imposing that and you will solve your limit. Some comments: the relevant limit that must be equal to the area of the circle is given by $$ r^2\pi\lim_{n\to\infty}\frac{\sin(2\pi/n)}{2\pi/n}. $$ Your limit is recover using the substitution $2\pi/n\mapsto x$ ($\infty\mapsto0$). The argoument of the limit follow by considering a subdivision of the $n$-gone in $n$ triangles that have the hipotenuse equal to the radius of the circle. The area of the triangle is found to be $$ A_T=r^22\frac{\cos(\pi/n)\sin(\pi/n)}{2} $$ that is equal to $$ r^2\frac{\sin(2\pi/n)}{2} $$ by the duplication formula. Clearly the area of circle $A_\mathcal{C}=\pi r^2$ is the limit of $nA_T$ when $n\to\infty$. Remember that in order to obtain the correct limit you have to multiply by $1$ written as $\pi/\pi$ and writing $n$ as $\frac{1}{1/n}$.

yngabl
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  • "The limit of the area as $n\to\infty$ must be equal to the area of the circle $\pi r^2$." This is one of those facts that are obvious to everyone. Yet, how do you prove this? – Ennar Nov 24 '17 at 15:28
  • @Ennar given that the OP is aware of the limit concept it must be at high school, then he is aware of what the area of the circle is.. – yngabl Nov 24 '17 at 15:30
  • I'm assuming we are all aware of everything here. I'm just kindly asking how exactly do you prove that area of regular $n$-gon with fixed radius $r$ tends to the area of a circle with radius $r$ when $n$ tends to infinity. – Ennar Nov 24 '17 at 15:32
  • @Ennar one can think the circle as a regular polygon with an infinite number of sides – yngabl Nov 24 '17 at 15:34
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    I can think of regular polygon as a funny looking duck. That's not a proof. In what topology does regular $n$-gon tend to a circle so we can think like that? And how does that relate in any way to their areas? – Ennar Nov 24 '17 at 15:35
  • @Ennar you can think whatever you want.. – yngabl Nov 24 '17 at 15:36
  • I think you are missing the point. The burden of the proof lies on you, not on me. – Ennar Nov 24 '17 at 15:37
  • I am not asking you to explain to me what you wrote, I understand everything perfectly. I'm asking you for a formal proof of "The limit of the area as $n\to\infty$ must be equal to the area of the circle $\pi r^2$." – Ennar Nov 24 '17 at 15:43
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I might be wrong, but I wager that starting from geometric definition of sine, what you are asking is impossible. The reasoning is simple. Although very easy to visualize, we actually have no clue what the value of $\sin x$ is exactly, except for some special $x$'s. It is inevitable that we approximate $\sin x$ the best we can and apply squeeze theorem to that.

Note that starting with definition $\sin x = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$, there is no squeezing needed, but then the problem is trivial since by definition $\lim_{x\to 0}\frac{\sin x}{x} = \sin'(0)$.

Ennar
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