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