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Question Number 134618 by Abdoulaye last updated on 05/Mar/21

  prove that cos (x) does not admit a  limt in +∞?

$$ \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{cos}\:\left(\mathrm{x}\right)\:\mathrm{does}\:\mathrm{not}\:\mathrm{admit}\:\mathrm{a} \\ $$$$\mathrm{limt}\:\mathrm{in}\:+\infty? \\ $$

Answered by Olaf last updated on 05/Mar/21

Let x_n  = cosn  Let y_n  = cos(n−1)  Let z_n  = cos(n+1)  Let t_n  = cos(2n)  Suppose now that x_n  admits a limit l  in +∞ :  lim_(n→∞)  x_n  = l  In this case we have too :  lim_(n→∞)  y_n  = lim_(n→∞)  z_n  = lim_(n→∞)  t_n  = l  But y_n +z_n  = 2x_n cos(1)  ⇒ l+l = 2l.cos(1) ⇒ l = 0 (1)  And t_n  = 2x_n ^2 −1  ⇒ l = 2l^2 −1 ⇒ l = −(1/2) or +1 (2)    (1) and (2) are not compatible !

$$\mathrm{Let}\:{x}_{{n}} \:=\:\mathrm{cos}{n} \\ $$$$\mathrm{Let}\:{y}_{{n}} \:=\:\mathrm{cos}\left({n}−\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{z}_{{n}} \:=\:\mathrm{cos}\left({n}+\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{t}_{{n}} \:=\:\mathrm{cos}\left(\mathrm{2}{n}\right) \\ $$$$\mathrm{Suppose}\:\mathrm{now}\:\mathrm{that}\:{x}_{{n}} \:\mathrm{admits}\:\mathrm{a}\:\mathrm{limit}\:{l} \\ $$$$\mathrm{in}\:+\infty\:: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{x}_{{n}} \:=\:{l} \\ $$$$\mathrm{In}\:\mathrm{this}\:\mathrm{case}\:\mathrm{we}\:\mathrm{have}\:\mathrm{too}\:: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{y}_{{n}} \:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{z}_{{n}} \:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{t}_{{n}} \:=\:{l} \\ $$$$\mathrm{But}\:{y}_{{n}} +{z}_{{n}} \:=\:\mathrm{2}{x}_{{n}} \mathrm{cos}\left(\mathrm{1}\right) \\ $$$$\Rightarrow\:{l}+{l}\:=\:\mathrm{2}{l}.\mathrm{cos}\left(\mathrm{1}\right)\:\Rightarrow\:{l}\:=\:\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{And}\:{t}_{{n}} \:=\:\mathrm{2}{x}_{{n}} ^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\:{l}\:=\:\mathrm{2}{l}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow\:{l}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{or}\:+\mathrm{1}\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right)\:\mathrm{are}\:\mathrm{not}\:\mathrm{compatible}\:! \\ $$

Answered by mathmax by abdo last updated on 06/Mar/21

if this limit exist for all sequence u_n /lim u_n =+∞  lim cos(u_n ) are the same  but for u_n =nπ  cos(u_n )=(−1)^n  and this sequence is not convergent...

$$\mathrm{if}\:\mathrm{this}\:\mathrm{limit}\:\mathrm{exist}\:\mathrm{for}\:\mathrm{all}\:\mathrm{sequence}\:\mathrm{u}_{\mathrm{n}} /\mathrm{lim}\:\mathrm{u}_{\mathrm{n}} =+\infty \\ $$$$\mathrm{lim}\:\mathrm{cos}\left(\mathrm{u}_{\mathrm{n}} \right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{same}\:\:\mathrm{but}\:\mathrm{for}\:\mathrm{u}_{\mathrm{n}} =\mathrm{n}\pi \\ $$$$\mathrm{cos}\left(\mathrm{u}_{\mathrm{n}} \right)=\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{and}\:\mathrm{this}\:\mathrm{sequence}\:\mathrm{is}\:\mathrm{not}\:\mathrm{convergent}... \\ $$

Commented by Abdoulaye last updated on 06/Mar/21

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by mathmax by abdo last updated on 06/Mar/21

you are welcome sir

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir} \\ $$

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