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My Teacher wants us to argue that this equation below is true. I am just curious how I would actually explain this? I mean I understand how its true but how can I argue this?

$\lim_{\theta\rightarrow 0^+}\frac{\cos\frac{1}{\theta}\sin\theta-\theta\cos\frac{1}{\theta}}{\theta} = 0$

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    @soniccool: I don't understand "u understand how its true": how do you understand that? Also, explaining a little of what you know about limits, and what you've tried already, would be useful.2012-06-25

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$ \lim_{\theta \rightarrow 0^+} \cos \left ( 1 \over \theta \right ) \left ( \sin \theta - \theta \over \theta \right) $ Let, $ \frac{1}{\theta} = \phi $, then we have for all real values of $ \phi $ $ -1 \leq \left ( \cos \phi \right ) \leq 1 $

EDIT:: taking comments into consideration $ - \left| \dfrac{\sin \theta}{\theta} - 1 \right| \le \cos(1/\theta) \left(\dfrac{\sin \theta}{\theta} - 1 \right) \le \left|\dfrac{\sin \theta}{\theta} - 1 \right| $ $ -1 \times 0 \leq \lim_{\phi \rightarrow \infty } \cos \phi \times \lim_{\theta \rightarrow 0 } \left ( \frac{\sin \theta} \theta - 1 \right ) \leq 1 \times 0 $ $ \text{Or, using squeeze theorem, we have, } \lim_{\theta \rightarrow 0^+} \cos \left ( 1 \over \theta \right ) \left ( \sin \theta - \theta \over \theta \right) = 0 $

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    @RobertIsrael thank you ...2012-06-25