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

 1)lim_(x→0) sin((1/x))=?  2) lim_(x→0) cos ((1/x))=?

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

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

1) limit dont exist let x_n =(1/((2n+1)(π/2)))  we have x_n →0 snd  sin(x_n )=sin((2n+1)(π/2))=sin(nπ +(π/2)) =cos(nπ)=(−1)^n  this sequence  hsvent any limit  2)no limit  for this function  take u_n =(1/(nπ))  (→0)  cos((1/u_n ))=cos(nπ)=(−1)^n  sequence no convergent...

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

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