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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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    It is best to leave the duplicated link in the body of the text, so that one can actually view the duplicated result!2012-08-31

2 Answers 2

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

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For sufficiently small positive values of $\theta$:

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

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

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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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    If they assume facts that are logically equivalent, then you can't use one to prove the other without descending into circular logic. My biggest issue with such proofs has always been $x\lt\tan x$ (which is equivalent to $\cos x\lt\frac{\sin x}{x}$). But upon further consideration, I realize that you could prove this fact by assuming that the boundary of a convex set is larger than the boundary of one of its convex subsets.2014-07-02