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Who was the first to prove $\lim_{x \to 0}\frac{\sin{x}}{x}=1$?

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I would guess the proof came "spontaneously" with the invention of limit or sine or its Taylor series, whichever comes first. It is like who was the first to prove $6+13 = 19$? –  Lord Soth Jun 4 '13 at 20:13
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Sandwich Theorem –  iostream007 Jun 4 '13 at 20:31
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This is an automatic consequence of the use of radian measure. The equivalent was well-known to Ptolemy a couple of thousand years ago, and undoubtedly earlier. –  André Nicolas Jun 4 '13 at 21:06
    
As André Nicolas says, the fact that $\sin x \approx x$ for very small $x$ is very ancient, and was probably even known to Archimedes. –  ShreevatsaR Aug 17 '13 at 7:16
    
Maybe it depends on what you consider a "proof". I think Euler wrote that when $x$ is an infinitely small non-zero number, then $\sin x=x$ and $\cos x=1$. –  Michael Hardy Oct 1 '13 at 17:49

4 Answers 4

up vote 1 down vote accepted

The first person to set out this limit explicitly, as far as I know, was Cauchy in his Cours d'analyse of 1821, pp. 66-67. See the English translation by Robert E. Bradley and C. Edward Sandifer, p. 45. Cauchy gives the limit again in his Resume des lecons (1823), lesson 1.

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The relation of $$ \sin x < x < \tan x, $$ when $x$ is small is implied by Taylor series. According to wikipedia, Brook Taylor was born in 1685.

According to wikipedia entry of pendulum:

Note that under the small-angle approximation, the period is independent of the amplitude $\theta_0$; this is the property of isochronism that Galileo discovered.

Galileo Galilei discovered this by small angle approximation: $$ \sin \theta \approx \theta \quad \text{ when } \theta \ll 1, $$ which dated back to early 1600s.

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You cannot use Taylor Series to prove such a limit. Taylor series use the derivative of $ \sin x $ and that limit is instrumental in computing the derivative. Rather, one can show the inequality using geometry of a circle and a triangle. –  Jon Claus Jun 4 '13 at 21:47
    
@JonClaus $$\sin x = x - \frac{x^3}{3!} + O(x^5) < x < \tan x = x + \frac{x^3}{3!} + O(x^5) $$ when $0<x\ll 1$. –  Shuhao Cao Jun 4 '13 at 21:56
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@ShuhaoCao The point is, how do you prove your approximations ($\sin x = x-\frac{x^3}{3!}+O(x^5)$, etc.) without knowing the derivative of $\sin x$? –  Steven Stadnicki Jun 4 '13 at 22:06
    
@StevenStadnicki Hmmm... Good point, I just meant to address the fact that the limit in the OP can be proved using Taylor like Lord Soth said earlier in the very first comment, but Galileo came up with small angle approximation using circles way earlier than that. –  Shuhao Cao Jun 4 '13 at 22:12
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@ShuhaoCao, you complete ignore my point. You cannot use the Taylor Series because that relies on the derivative of $ \sin x $ (by definition) which requires you to know the limit of $ \frac{\sin x}{x} $. Look at the post below to see why if you don't know. –  Jon Claus Jun 5 '13 at 2:36

My guess is that this identity, while perhaps not "proved" with the same rigor as we expect from proofs today, was known in antiquity. Certainly by the 10th century, when Islamic mathematicians were using all six trigonometric functions in essentially their modern form, since the identity can be deduced from a geometric argument. The general concept of a limit is also much older than the development of calculus or the limit's eventual rigorous definition, and the Squeeze (or Sandwich) theorem was known in some form to the ancient Greeks.

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GH Hardy "Pure Mathematics" 10th Edition p433 (Chapter IX,section 224) says:

The most natural way [of defining the trigonometrical functions] is to follow as closely as we can the procedure of the ordinary textbooks, translating the geometrical language which they employ into the language of analysis. We discussed this problem in [sect 163] and concluded that it involves one and only one serious difficulty. We have to show either that with any arc of a circle is associated a number which we call its length, or that with any sector of a circle is associated a number which we call its area.

He procedes by using area, which has been defined via integration. If you are interested in the foundational questions, there is some interesting discussion of the issues with various approaches. However, he does not discuss the first person to make this rigorous, and I would be interested to see other references. This does not answer the question of the limit, of course, but shows that if rigour is required some care is useful in the definition of $\sin x$ even before the limit is considered.

Hardy considers the limit in the question on page 182 Examples XXXVI 13, leaving the rigour till later.

This was too long for a comment.

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