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Question-194139




Question Number 194139 by cortano12 last updated on 28/Jun/23
Commented by MM42 last updated on 28/Jun/23
for  lim_(x→0)  ((x^2 +2cosx−2)/x^4 ) →hop  lim_(x→0)  ((2(x−sinx))/(4x^3 )) =(1/2)×(1/6)=(1/(12))  ⇒
$${for} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}^{\mathrm{2}} +\mathrm{2}{cosx}−\mathrm{2}}{{x}^{\mathrm{4}} }\:\rightarrow{hop} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{2}\left({x}−{sinx}\right)}{\mathrm{4}{x}^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\Rightarrow \\ $$
Answered by MM42 last updated on 28/Jun/23
hop→lim_(x→0)  ((2x+2sinx)/(4x^3 )) =+∞
$${hop}\rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{2}{x}+\mathrm{2}{sinx}}{\mathrm{4}{x}^{\mathrm{3}} }\:=+\infty \\ $$$$ \\ $$

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