Question Number 197336 by Huy last updated on 13/Sep/23
$${lim}_{{x}\rightarrow+\infty} \left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{cos}{x}\right)=? \\ $$
Answered by TheHoneyCat last updated on 13/Sep/23
$$\mathrm{This}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{undefined} \\ $$$$\left(\mathrm{because}\:\mathrm{lim}\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{0}\:\mathrm{and}\:\mathrm{lim}\:\mathrm{cos}{x}\:\mathrm{is}\:\mathrm{undefined}\right) \\ $$
Answered by MM42 last updated on 13/Sep/23
$${let}\:\:{f}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+{cosx} \\ $$$${for}\:\:{a}_{{n}} =\mathrm{2}{n}\pi\Rightarrow{lim}_{{n}\rightarrow+\infty} {f}\left({a}_{{n}} \right)={cos}\mathrm{2}{n}\pi=\mathrm{1} \\ $$$${for}\:\:{b}_{{n}} =\mathrm{2}{n}\pi+\pi\Rightarrow{lim}_{{n}\rightarrow+\infty} {f}\left({b}_{{n}} \right)={cos}\pi=−\mathrm{1} \\ $$$$\Rightarrow\:{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)={not}\:\:{exist} \\ $$$$ \\ $$