CalculusEasyLimits and Continuity ShareLimits and Continuity · Easy problemUsing the special limit limu→0sinuu=1\displaystyle\lim_{u\to0}\dfrac{\sin u}{u}=1u→0limusinu=1, evaluate limx→0xsin4x\displaystyle\lim_{x\to 0}\dfrac{x}{\sin 4x}x→0limsin4xx.Check Show answer Walk through it step by step Practice more like this Review this concept