Solution :
limx→0 2(sin(2+x)−sin2)+xsin(2+x)x
= limx→0(2.2.cos(2+x2)sinx2x + sin(2+x))
= limx→02cos(2+x2)sinx2x2 + limx→0sin(2+x)
= 2cos2 + sin2
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