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Question Number 141932 by mathmax by abdo last updated on 24/May/21
calculate lim_(x→0)  ((sin(sin(sinx))+1−cos(x^2 ))/x^3 )
$$\mathrm{calculate}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{sin}\left(\mathrm{sin}\left(\mathrm{sinx}\right)\right)+\mathrm{1}−\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{3}} } \\ $$
Answered by TheSupreme last updated on 24/May/21
sin(sin(sin(x)))+1−cos(x^2 )→x+(x^4 /2)  so lim =∞
$${sin}\left({sin}\left({sin}\left({x}\right)\right)\right)+\mathrm{1}−{cos}\left({x}^{\mathrm{2}} \right)\rightarrow{x}+\frac{{x}^{\mathrm{4}} }{\mathrm{2}} \\ $$$${so}\:{lim}\:=\infty \\ $$
Answered by mathmax by abdo last updated on 26/May/21
sinx ∼x ⇒sin(sin(sinx)∼ x⇒f(x)= ((sin(sin(sinx)+1−cosx^2 )/x^3 )  ∼((x+1−cos(x^2 ))/x^3 ) =(1/x^2 )+((1−cos(x^2 ))/x^2 )∼(1/x^2 ) +(x^4 /(2x^2 ))=(1/x^2 )+(x^2 /2) ⇒  lim_(x→0)   f(x)=+∞
$$\mathrm{sinx}\:\sim\mathrm{x}\:\Rightarrow\mathrm{sin}\left(\mathrm{sin}\left(\mathrm{sinx}\right)\sim\:\mathrm{x}\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\:\frac{\mathrm{sin}\left(\mathrm{sin}\left(\mathrm{sinx}\right)+\mathrm{1}−\mathrm{cosx}^{\mathrm{2}} \right.}{\mathrm{x}^{\mathrm{3}} }\right. \\ $$$$\sim\frac{\mathrm{x}+\mathrm{1}−\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} }\sim\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{2x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\mathrm{f}\left(\mathrm{x}\right)=+\infty \\ $$$$ \\ $$

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