PrecalculusSuperchallengeIntroduction to Limits ShareIntroduction to Limits · Superchallenge problemUsing limu→0sinuu=1\displaystyle\lim_{u \to 0}\frac{\sin u}{u} = 1u→0limusinu=1, evaluate limx→π/31−2cosxsin(x−π3)\displaystyle\lim_{x \to \pi/3} \frac{1 - 2\cos x}{\sin\left(x - \dfrac{\pi}{3}\right)}x→π/3limsin(x−3π)1−2cosx.Check Show answer Walk through it step by step Practice more like this Review this concept