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Question Number 91930 by Ozoda last updated on 03/May/20

lim_(x→0) cos (1/x)=

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}cos}\:\frac{\mathrm{1}}{{x}}= \\ $$

Commented by mr W last updated on 03/May/20

when x→0, (1/x)→∞, cos (1/x) doesn′t  approach a certain value, therefore  lim_(x→0)  cos (1/x) doesn′t exist.

$${when}\:{x}\rightarrow\mathrm{0},\:\frac{\mathrm{1}}{{x}}\rightarrow\infty,\:\mathrm{cos}\:\frac{\mathrm{1}}{{x}}\:{doesn}'{t} \\ $$$${approach}\:{a}\:{certain}\:{value},\:{therefore} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{cos}\:\frac{\mathrm{1}}{{x}}\:{doesn}'{t}\:{exist}. \\ $$

Commented by turbo msup by abdo last updated on 03/May/20

if this limit exist we must hsve  of all sequence (x_n ) →0 limcos(x_n )  exist let x_n =(1/(nπ))  we have x_n →0 sndd  cos(x_n )=cos(nπ) =(−1)^n  this  sequence have nt any limit →  lim_(x→0)  cosx dont exist..

$${if}\:{this}\:{limit}\:{exist}\:{we}\:{must}\:{hsve} \\ $$$${of}\:{all}\:{sequence}\:\left({x}_{{n}} \right)\:\rightarrow\mathrm{0}\:{limcos}\left({x}_{{n}} \right) \\ $$$${exist}\:{let}\:{x}_{{n}} =\frac{\mathrm{1}}{{n}\pi}\:\:{we}\:{have}\:{x}_{{n}} \rightarrow\mathrm{0}\:{sndd} \\ $$$${cos}\left({x}_{{n}} \right)={cos}\left({n}\pi\right)\:=\left(−\mathrm{1}\right)^{{n}} \:{this} \\ $$$${sequence}\:{have}\:{nt}\:{any}\:{limit}\:\rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:{cosx}\:{dont}\:{exist}.. \\ $$

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