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Question Number 40095 by maxmathsup by imad last updated on 15/Jul/18
let f(x) =cos(x)cos((1/x))  is f  have a limit at point 0?
$${let}\:{f}\left({x}\right)\:={cos}\left({x}\right){cos}\left(\frac{\mathrm{1}}{{x}}\right)\:\:{is}\:{f}\:\:{have}\:{a}\:{limit}\:{at}\:{point}\:\mathrm{0}? \\ $$$$ \\ $$
Answered by math khazana by abdo last updated on 26/Jul/18
let x_n = (2/(nπ))  we have lim_(n→+∞)  x_n =0 and  f(x_n ) =cos((2/(nπ)))cos(((nπ)/2)) ⇒  f(x_(2n) ) =cos((2/(2nπ)))cos(((2nπ)/2))=cos((1/(nπ)))(−1)^n   and (−1)^n  dont have any limit and  that prove  that f haven t any limit at point 0
$${let}\:{x}_{{n}} =\:\frac{\mathrm{2}}{{n}\pi}\:\:{we}\:{have}\:{lim}_{{n}\rightarrow+\infty} \:{x}_{{n}} =\mathrm{0}\:{and} \\ $$$${f}\left({x}_{{n}} \right)\:={cos}\left(\frac{\mathrm{2}}{{n}\pi}\right){cos}\left(\frac{{n}\pi}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${f}\left({x}_{\mathrm{2}{n}} \right)\:={cos}\left(\frac{\mathrm{2}}{\mathrm{2}{n}\pi}\right){cos}\left(\frac{\mathrm{2}{n}\pi}{\mathrm{2}}\right)={cos}\left(\frac{\mathrm{1}}{{n}\pi}\right)\left(−\mathrm{1}\right)^{{n}} \\ $$$${and}\:\left(−\mathrm{1}\right)^{{n}} \:{dont}\:{have}\:{any}\:{limit}\:{and}\:\:{that}\:{prove} \\ $$$${that}\:{f}\:{haven}\:{t}\:{any}\:{limit}\:{at}\:{point}\:\mathrm{0} \\ $$

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