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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)

$${find}\:{a}\:{equivalent}\:{to}\:{f}\left({x}\right)={cos}\left({sinx}\right)\:{for}\:{x}\in{v}\left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{a}\:{equivalent}\:{to}\:{g}\left({x}\right)=\:{tan}\left(\frac{\pi}{\mathrm{2}{x}+\mathrm{1}}\right)\:\left({x}\rightarrow\mathrm{0}\right) \\ $$

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)

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{sinx}\:\sim{x}\:\Rightarrow{cos}\left({sinx}\right)\sim{cosx}\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left({x}\rightarrow\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:\:\frac{\pi}{\mathrm{2}{x}+\mathrm{1}}\:=\frac{\pi}{\mathrm{1}+\mathrm{2}{x}}\:\:\sim\pi\left(\mathrm{1}−\mathrm{2}{x}\right)\:\Rightarrow \\ $$$${tan}\left(\frac{\pi}{\mathrm{2}{x}+\mathrm{1}}\right)\:\sim{tan}\left(\pi\left(\mathrm{1}−\mathrm{2}{x}\right)\right)\sim−{tan}\left(\mathrm{2}\pi{x}\right)\sim−\mathrm{2}\pi{x}\:\Rightarrow \\ $$$${g}\left({x}\right)\sim−\mathrm{2}\pi{x}\:\left({x}\rightarrow\mathrm{0}\right) \\ $$

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