lim_{α \to 0} \frac{\sin α}{α} = \frac{[\sin α]^{'}}{α^{'}} = \frac{\cos 0}{1} = 1

${\lim_{\alpha \rightarrow 0} \frac{\sin \alpha}{ \alpha} = \frac{[\sin \alpha]'}{\alpha'} = \frac{\cos 0}{1} = 1}$

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