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CalculusEasyLimits and Continuity

Limits and Continuity · Easy problem

Using the special limit lim⁡u→0sin⁡uu=1\displaystyle\lim_{u\to0}\dfrac{\sin u}{u}=1, evaluate lim⁡x→0xsin⁡4x\displaystyle\lim_{x\to 0}\dfrac{x}{\sin 4x}.