# Evaluate the limit: #lim_(theta-> 0)(1-cos theta)/(sin 2theta)#?

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The limit is 0.

This limit can be evaluated in (at least) two ways. The first is by using l'Hopital's Rule, which states

By using the identities of Pythagore, we have

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To evaluate the limit ( \lim_{\theta \to 0} \frac{1 - \cos(\theta)}{\sin(2\theta)} ), apply L'Hôpital's Rule:

[ \lim_{\theta \to 0} \frac{1 - \cos(\theta)}{\sin(2\theta)} = \lim_{\theta \to 0} \frac{d/d\theta(1 - \cos(\theta))}{d/d\theta(\sin(2\theta))} ]

[ = \lim_{\theta \to 0} \frac{\sin(\theta)}{2\cos(2\theta)} ]

Now, plug in ( \theta = 0 ) to get:

[ \frac{\sin(0)}{2\cos(0)} = \frac{0}{2 \cdot 1} = 0 ]

So, ( \lim_{\theta \to 0} \frac{1 - \cos(\theta)}{\sin(2\theta)} = 0 ).

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