CalculusEasyLimits and Continuity ShareLimits and Continuity · Easy problemUse the special trigonometric limit limu→0sinuu=1\displaystyle\lim_{u\to0}\dfrac{\sin u}{u}=1u→0limusinu=1 to evaluate limx→0sin7xx\displaystyle\lim_{x\to 0}\dfrac{\sin 7x}{x}x→0limxsin7x.Check Show answer Walk through it step by step Practice more like this Review this concept