PrecalculusChallengeIntroduction to Limits ShareIntroduction to Limits · Challenge problemUsing the fact that limx→0sinxx=1\displaystyle\lim_{x \to 0}\frac{\sin x}{x} = 1x→0limxsinx=1, evaluate limx→01−cos2xxsinx\displaystyle\lim_{x \to 0} \frac{1 - \cos 2x}{x\sin x}x→0limxsinx1−cos2x.Check Show answer Walk through it step by step Practice more like this Review this concept