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Question Number 197336 by Huy last updated on 13/Sep/23

$${lim}_{x\to +\mathrm{\infty }}(\frac{1}{x^2}+\cos x)=?$$

Answered by TheHoneyCat last updated on 13/Sep/23

$$\mathrm{This}\mathrm{limit}\mathrm{is}\mathrm{undefined}$$$$(\mathrm{because}\lim \frac{1}{x^2}=0\mathrm{and}\lim \cos x\mathrm{is}\mathrm{undefined})$$

Answered by MM42 last updated on 13/Sep/23

$$letf\left(x\right)=\frac{1}{x^2}+cosx$$$$for{a}_n=2n\pi \Rightarrow {lim}_{n\to +\mathrm{\infty }}f\left(a_n\right)=cos2n\pi =1$$$$for{b}_n=2n\pi +\pi \Rightarrow {lim}_{n\to +\mathrm{\infty }}f\left(b_n\right)=cos\pi =-1$$$$\Rightarrow {lim}_{x\to +\mathrm{\infty }}f\left(x\right)=notexist$$$$$$

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