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PrecalculusChallengeIntroduction to Limits

Introduction to Limits · Challenge problem

Using the fact that lim⁡x→0sin⁡xx=1\displaystyle\lim_{x \to 0}\frac{\sin x}{x} = 1, evaluate lim⁡x→01−cos⁡2xxsin⁡x\displaystyle\lim_{x \to 0} \frac{1 - \cos 2x}{x\sin x}.