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Question Number 110647 by Khalmohmmad last updated on 29/Aug/20

$$1)\underset{x\to 0}{\lim sin}\left(\frac{1}{\mathrm{x}}\right)=?$$$$2)\underset{x\to 0}{\lim cos}\left(\frac{1}{\mathrm{x}}\right)=?$$$$$$

Answered by mathmax by abdo last updated on 30/Aug/20

$$1)\mathrm{limit}\mathrm{dont}\mathrm{exist}\mathrm{let}{\mathrm{x}}_{\mathrm{n}}=\frac{1}{(2\mathrm{n}+1)\frac{\pi }{2}}\mathrm{we}\mathrm{have}{\mathrm{x}}_{\mathrm{n}}\to 0\mathrm{snd}$$$$\sin \left({\mathrm{x}}_{\mathrm{n}}\right)=\sin \left((2\mathrm{n}+1)\frac{\pi }{2}\right)=\sin (\mathrm{n}\pi +\frac{\pi }{2})=\cos \left(\mathrm{n}\pi \right)={(-1)}^{\mathrm{n}}\mathrm{this}\mathrm{sequence}$$$$\mathrm{hsvent}\mathrm{any}\mathrm{limit}$$$$2)\mathrm{no}\mathrm{limit}\mathrm{for}\mathrm{this}\mathrm{function}\mathrm{take}{\mathrm{u}}_{\mathrm{n}}=\frac{1}{\mathrm{n}\pi }(\to 0)$$$$\cos \left(\frac{1}{{\mathrm{u}}_{\mathrm{n}}}\right)=\cos \left(\mathrm{n}\pi \right)={(-1)}^{\mathrm{n}}\mathrm{sequence}\mathrm{no}\mathrm{convergent}...$$

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