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→∞x(π2−arctanx)\displaystyle\lim_{x \to \infty} x\left(\frac{\pi}{2} - \arctan x\right)x→∞limx(2π−arctanx).Check Show answer Walk through it step by step Practice more like this Review this concept