Why the limit of $\frac{\sin(x)}{x}$ as $x$ approaches 0 is 1?

Question

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.

Sure, but i need to know the origin of this theorem, without l'hopital's rule. :) — egarro, Aug 24 '14 at 02:55
@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. — Michael Albanese, Aug 24 '14 at 02:56
@msteve What i need is the formal proof of this theorem. But actually i'm a noob making proofs. :( How can i do it with Taylor expansion? — egarro, Aug 24 '14 at 03:00
@JohnD I don't understand in the video why there's a point with (1,tan(x))? — egarro, Aug 24 '14 at 03:22
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 $\sin\theta\over \cos\theta$. — JohnD, Aug 24 '14 at 04:00
You might be interested in looking at the proofs here: https://proofwiki.org/wiki/Limit_of_Sine_of_X_over_X — David H, Aug 24 '14 at 04:17
Note that you need a rigorous definition of $\sin(x)$ before you can hope to have a rigorous proof that $\lim_{x \to 0} \sin(x)/x = 1$. — , Aug 24 '14 at 04:25
@Hurkyl And what can you suggest me for a rigorous definition of sin(x)? — egarro, Aug 24 '14 at 04:33
@egarro: There are a number of approaches that people take. A common one is to take what we know to be the Taylor series for $\sin(x)$ about zero, and define $\sin(x)$ to be the value of the Taylor series. A rigorous geometric definition of $\sin(x)$ can be done, but it has a lot of annoying details you have to work through first (e.g. rigorously define things like "arclength" and "angle" and "radian"). Another common one is by its differential equation: $f = \sin$ is the unique solution to $f''(x) = -f(x)$, $f(0) = 0$ and $f'(0)=1$. — , Aug 24 '14 at 04:37
@Hurkyl can you recommend a book for learn more about this topics? — egarro, Aug 24 '14 at 04:42
@egarro One neat way to define the sine and cosine functions is as the solutions to a particular set of functional equations. Part of the definition requires the sine function to satisfy a certain basic inequality, which you can then apply the squeeze theorem to. — David H, Aug 24 '14 at 04:44
Plot on the same graph $y=\sin(x)$ and $y=x$ for $0 \leq x \leq 1$. You will see how close the curves are. — Claude Leibovici, Aug 24 '14 at 04:44
@egarro: I am unable to give a recommendation. Consider looking at some of the more rigorous introductory calculus texts or maybe an advanced calculus text. Or possibly introductory complex analysis texts, as they have to introduce things like $\sin(z)$ for complex values of $z$ which is expected to be a novel thing to its audience and requiring careful justification. — , Aug 24 '14 at 04:49

user164587 · Answer 1 · 2014-09-07T03:26:26.537

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 $$\lim_{\alpha \to 0} \frac{\sin \alpha}{\alpha} = 1$$

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... — egarro, Aug 24 '14 at 03:26
@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 $\sin(x)$. — jxnh, Aug 24 '14 at 03:28

score 3 · Answer 2 · edited Sep 08 '15 at 04:18

3

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 $.

edited Sep 08 '15 at 04:18

Berrick Caleb Fillmore

5,269

answered Aug 24 '14 at 03:34

narendra-choudhary

770

1

you think that using taylor series it will be right for a formal proof? you know for a test or something like that... ;) – egarro Aug 24 '14 at 03:43
4

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. – voldemort Aug 24 '14 at 03:46
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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. – narendra-choudhary Aug 24 '14 at 03:56
@user3679857 Thanks for your help ;) – egarro Aug 24 '14 at 03:58
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Don't you need derivative of Sinx in Taylor expansion and you need the same limit to find derivative of Sinx? – Janaka Rodrigo Aug 06 '22 at 03:10

Why the limit of $\frac{\sin(x)}{x}$ as $x$ approaches 0 is 1?

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