Respuesta :

dalendrk
[tex]We\ know:\lim\limits_{x\to0}\dfrac{\sin x}{x}=1\\-------------------\\\lim\limits_{x\to0}\dfrac{\sin4x}{2x}=\lim\limits_{x\to0}\dfrac{2\sin4x}{2\cdot2x}=\lim\limits_{x\to0}2\cdot\dfrac{\sin4x}{4x}=2\lim\limits_{x\to0}\dfrac{\sin4x}{4x}=(*)\\\\subtitute\ 4x=t;\ x\to0\ then\ 4x\to0\ then\ t\to 0\\\\(*)=2\lim\limits_{t\to0}\dfrac{\sin t}{t}=2\cdot1=2[/tex]

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