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Question Number 40091 by maxmathsup by imad last updated on 15/Jul/18

find a equivalent to f(x)=cos(sinx) for x∈v(0)  2) find a equivalent to g(x)= tan((π/(2x+1))) (x→0)

findaequivalenttof(x)=cos(sinx)forxv(0)2)findaequivalenttog(x)=tan(π2x+1)(x0)

Commented by math khazana by abdo last updated on 26/Jul/18

1) we have sinx ∼x ⇒cos(sinx)∼cosx∼1−(x^2 /2)  ⇒f(x)∼1−(x^2 /2)(x→0)  2) we have  (π/(2x+1)) =(π/(1+2x))  ∼π(1−2x) ⇒  tan((π/(2x+1))) ∼tan(π(1−2x))∼−tan(2πx)∼−2πx ⇒  g(x)∼−2πx (x→0)

1)wehavesinxxcos(sinx)cosx1x22f(x)1x22(x0)2)wehaveπ2x+1=π1+2xπ(12x)tan(π2x+1)tan(π(12x))tan(2πx)2πxg(x)2πx(x0)

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