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Proving $\lim_{\theta \to 0}{\frac{\sin \theta}{\theta}}=1$ using $\frac{1}{2}r^2(\theta-\sin \theta)$

How can one prove that $$\lim_{\theta \to 0} \frac{\sin\theta}{\theta}=1$$

$\sin(0) = 0$

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marked as duplicate by Sasha, DonAntonio, William, David Mitra, Henry T. Horton Aug 31 '12 at 3:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
It is best to leave the duplicated link in the body of the text, so that one can actually view the duplicated result! –  JavaMan Aug 31 '12 at 7:12

3 Answers 3

For sufficiently small positive values of $\theta$:

$$\sin \theta \leq \theta \leq \tan\theta$$

This can be shown geometrically (picture is from here):

enter image description here

The area of $\Delta OAC$ is $\frac 12 \cdot 1 \cdot \sin x=\frac 12\sin x$. The area of the circular sector $AOC$ is $\frac x2$. The area of triangle $\Delta OAB$ is $\frac12\cdot 1\cdot \tan x=\frac 12\tan x$. Noting that these shapes are contained within each other (and rewriting the variable $x$ as $\theta$ for consistency), we write

$$\frac 12\sin \theta\leq\frac \theta2\leq\frac 12\tan \theta$$ $$1\leq\frac{\theta}{\sin\theta}\leq\frac 1 {\cos\theta}$$ $$\cos\theta\leq\frac{\sin\theta}{\theta}\leq1$$

Take $\lim_{\theta\to 0^+}\cos\theta$ and apply the squeeze theorem. Just flip the inequalities around in the first line for the negative case. When you go to the second line by dividing by $\sin\theta$, which is negative then $\theta$ is negative, the rest of the argument proceeds the exact same way.

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How does one show that $\sin(\theta)\le\theta\le\tan(\theta)$? –  robjohn Mar 27 '13 at 17:44
    
There's a great treatment of a geometric proof by areas here. Should I update my answer with more information? –  Robert Mastragostino Mar 28 '13 at 12:03
    
That looks very similar to my answer. Since this is a key inequality, something more might be said. –  robjohn Mar 28 '13 at 14:55
    
how does one show that the area of a circle is $\pi r^2$ without assuming that $\lim\limits_{x\to 0}\frac{\sin x}{x} = 1$? see math.stackexchange.com/questions/605915/… for a similar answer. –  John Joy Jul 1 at 15:23
    
@JohnJoy what's your definition of $\pi$ to start with? There are plenty that don't need to make any reference to trig functions whatsoever. They might assume facts that end up being logically equivalent to $\lim_{x\to 0} \sin x/x=1$, but I don't see how it's inevitably circular. –  Robert Mastragostino Jul 1 at 22:09

There are many ways. The easiest way, I believe, is by L'Hôpital's rule.

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I've always disliked this justification, because it is almost always the case that one shows that the derivative of $\sin$ is $\cos$ using the limit asked about, meaning that this is (likely) circular. –  mixedmath Aug 31 '12 at 3:45
    
if you dislike this, try power series version : $sin(h)=h-h^3/3!+h^5/5!+...$, then $sin(h)/h=1-h^2/3!+h^4/5!+...$, and the limit when h goes to zero then is 1. –  sara Aug 31 '12 at 5:07
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@Solmaz: Again this is circular. The Taylor series of $sin x$ requires one to know derivatives of $\sin x$. –  JavaMan Aug 31 '12 at 7:10
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@JavaMan: I believe that sara is using the power series as the definition for $\sin$ and $\cos$. –  robjohn Mar 27 '13 at 17:45

$\lim_{\theta \rightarrow 0} \frac{\sin(\theta)}{\theta} = \lim_{\theta \rightarrow 0} \cos(\theta) = 1$ by L'hopital's rule.

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