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I need a rigorous proof that verify why the limit of $\dfrac{\sin(x)}{x}$ as $x$ approaches $0$ is $1$. I tried before but i do not know how start this proof. I would appreciate if somebody help me. Thanks.

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marked as duplicate by Andrés E. Caicedo, Jyrki Lahtonen, Claude Leibovici, Tunk-Fey, user98602 Aug 24 '14 at 6:57

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    $\begingroup$ You could use the Taylor expansion of sin about zero? $\endgroup$ – msteve Aug 24 '14 at 2:56
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    $\begingroup$ @AlexG.: Using L'Hôpital's rule would be circular. In order to prove that the derivative of $\sin$ is $\cos$, you need to know the limit in the question. $\endgroup$ – Michael Albanese Aug 24 '14 at 2:56
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    $\begingroup$ Read Robjohn's answer here $\endgroup$ – minibuffer Aug 24 '14 at 3:08
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    $\begingroup$ The big triangle is a right triangle, and $\tan\theta={\text{opposite}\over \text{adjacent}}$, but the adjacent side is already 1, making the length of that vertical side $\tan\theta$, which he writes as $\sin\theta\over \cos\theta$. $\endgroup$ – JohnD Aug 24 '14 at 4:00
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    $\begingroup$ Note that you need a rigorous definition of $\sin(x)$ before you can hope to have a rigorous proof that $\lim_{x \to 0} \sin(x)/x = 1$. $\endgroup$ – Hurkyl Aug 24 '14 at 4:25
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Here's a proof by the squeeze theorem.

Consider a unit circle as in the diagram below. enter image description here

The right-angled triangle ABC has hypotenuse 1 because it is a radius of the unit circle. So BC has length $\sin \alpha$.

Similarly, the right-angled triangle ADE has adjacent 1 because it is a radius of the unit circle. So DE has length $\tan \alpha$.

Then the triangle ABE has area $\frac12 b \times h = \frac12 \sin \alpha$. The sector ABE has area $\frac12 \alpha$. And the triangle ADE has area $\frac12 b \times h = \frac12 \tan \alpha$.

And we can clearly see that $$\frac12 \sin \alpha \lt \frac12 \alpha \lt \frac12 \tan \alpha$$ Dividing by $\frac12 \sin \alpha$ and taking the reciprocals gets us $$1 \gt \frac{\sin \alpha}{\alpha} \gt \cos \alpha$$ Then, taking the limit as $\alpha \to 0^+$ we have $$1 \ge \lim_{\alpha \to 0^+} \frac{\sin \alpha}{\alpha} \ge 1 \implies \lim_{\alpha \to 0^+} \frac{\sin \alpha}{\alpha} = 1$$ Then setting $\beta = -\alpha$ we get $$\lim_{\beta \to 0^-} \frac{\sin \beta}{\beta} = 1$$ And thus $$\lim_{\alpha \to 0} \frac{\sin \alpha}{\alpha} = 1$$

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  • $\begingroup$ Sorry but i didn't get the video's explanation, why there's a side of the triangle named as sin(x), and other as tan(x)? They actually are sides, not trigonometric functions as i know... $\endgroup$ – egarro Aug 24 '14 at 3:26
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    $\begingroup$ @egarro, the trigonometric functions refer to triangles. If you have a right triangle with appropriate angle $x$ and hypotenuse $1$ then the opposite side will by definition be of length $\sin(x)$. $\endgroup$ – JHance Aug 24 '14 at 3:28
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You can expand $ \sin(x) $ using a Taylor series: $$ \sin(x) \approx x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \cdots + (-1)^{n} \cdot \frac{x^{2 n + 1}}{(2 n + 1)!}. $$ Hence, $$ \frac{\sin(x)}{x} \approx 1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - \frac{x^{6}}{7!} + \cdots + (-1)^{n} \cdot \frac{x^{2 n}}{(2 n + 1)!}. $$ Therefore, as $ x $ tends to zero, $ \dfrac{\sin(x)}{x} $ tends to $ 1 $.

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    $\begingroup$ you think that using taylor series it will be right for a formal proof? you know for a test or something like that... ;) $\endgroup$ – egarro Aug 24 '14 at 3:43
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    $\begingroup$ Disagree with this proof. The individual terms go to $0$, but that does not guarantee that the sum goes to $0$ as well- one has to be careful. $\endgroup$ – voldemort Aug 24 '14 at 3:46
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    $\begingroup$ You can't calculate exact value of sin(x)/x for x=$0$. When you say x tends to $0$, you're already taking an approximation.So, we have to calculate the limit here.Taylor series gives very accurate approximation of sin(x), so it can be used to calculate limit. $\endgroup$ – narendra-choudhary Aug 24 '14 at 3:56
  • $\begingroup$ @user3679857 Thanks for your help ;) $\endgroup$ – egarro Aug 24 '14 at 3:58

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