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Question Number 84460 by jagoll last updated on 13/Mar/20
limx→0sin(2+x)−sin(2−x)x
Commented by jagoll last updated on 13/Mar/20
limx→0cos(2+x)+cos(2−x)1=2cos2
Commented by mathmax by abdo last updated on 13/Mar/20
letf(x)=sin(2+x)−sin(2−x)x⇒f(x)=sin2cosx+cos2sinx−(sin2cosx−cos2sinx)x=2cos2sinxx∼2cos2(x→0)⇒limx→0f(x)=2cos2
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