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Question Number 40091 by maxmathsup by imad last updated on 15/Jul/18
findaequivalenttof(x)=cos(sinx)forx∈v(0)2)findaequivalenttog(x)=tan(π2x+1)(x→0)
Commented by math khazana by abdo last updated on 26/Jul/18
1)wehavesinx∼x⇒cos(sinx)∼cosx∼1−x22⇒f(x)∼1−x22(x→0)2)wehaveπ2x+1=π1+2x∼π(1−2x)⇒tan(π2x+1)∼tan(π(1−2x))∼−tan(2πx)∼−2πx⇒g(x)∼−2πx(x→0)
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