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Question Number 34307 by prof Abdo imad last updated on 03/May/18

let give A_n = (((1 (α/n))),((−(α/n) 1)) ) calculate lim_(n→+∞) A_n ^n .

$${let}\:{give}\:{A}_{{n}} =\:\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\frac{\alpha}{{n}}}\\{−\frac{\alpha}{{n}}\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} ^{{n}} \:\:\:\:. \\ $$

Commented by math khazana by abdo last updated on 06/May/18

we have (√(1 +(α^2 /n^2 ))) =(1/n)(√(n^2 +α^2 )) ⇒ A_n =(√(1+(α^2 /n^2 ))) . ((( (1/(√(1+(α^2 /n^2 )))) (α/(n(√(1+(α^2 /n^2 ))))))),((−(α/(n(√(1+(α^2 /n^2 ))))) (1/(√(1+(α^2 /n^2 )))))) ) A_n =(√(1+(α^2 /n^2 ))) (((cosθ_ sinθ)),((−sinθ cosθ)) ) with cosθ = (1/(√(1+(α^2 /n^n )))) and sinθ = (α/(n.(√(1+(α^2 /n^2 ))))) ⇒ tanθ= (α/n) ⇒ θ=arctan((α/n)) we have A_n ^n = (1+(α^2 /n^2 ))^(n/2) . (((cos(nθ) sin(nθ))),((−sin(nθ) cos(nθ)) ) but (1+(α^2 /n^2 ))^(n/2) =e^((n/2)ln(1+(α^2 /n^2 )) ) ∼ e^(α^2 /(2n)) →1(n→+∞) n arctan((α/n)) →α so lim_(n→+∞) A_n ^n = (((cosα sinα)),((−sinα cosα)) )

$${we}\:{have}\:\sqrt{\mathrm{1}\:+\frac{\alpha^{\mathrm{2}} }{{n}^{\mathrm{2}} }}\:=\frac{\mathrm{1}}{{n}}\sqrt{{n}^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\:\:\Rightarrow \\ $$$${A}_{{n}} =\sqrt{\mathrm{1}+\frac{\alpha^{\mathrm{2}} }{{n}^{\mathrm{2}} }}\:\:.\begin{pmatrix}{\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\frac{\alpha^{\mathrm{2}} }{{n}^{\mathrm{2}} }}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\alpha}{{n}\sqrt{\mathrm{1}+\frac{\alpha^{\mathrm{2}} }{{n}^{\mathrm{2}} }}}}\\{−\frac{\alpha}{{n}\sqrt{\mathrm{1}+\frac{\alpha^{\mathrm{2}} }{{n}^{\mathrm{2}} }}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\frac{\alpha^{\mathrm{2}} }{{n}^{\mathrm{2}} }}}}\end{pmatrix} \\ $$$${A}_{{n}} =\sqrt{\mathrm{1}+\frac{\alpha^{\mathrm{2}} }{{n}^{\mathrm{2}} }}\:\:\begin{pmatrix}{{cos}\theta_{\:} \:\:\:\:\:\:\:\:\:\:\:{sin}\theta}\\{−{sin}\theta\:\:\:\:\:\:\:\:{cos}\theta}\end{pmatrix} \\ $$$${with}\:{cos}\theta\:\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\frac{\alpha^{\mathrm{2}} }{{n}^{{n}} }}}\:\:{and}\:{sin}\theta\:=\:\frac{\alpha}{{n}.\sqrt{\mathrm{1}+\frac{\alpha^{\mathrm{2}} }{{n}^{\mathrm{2}} }}}\:\Rightarrow \\ $$$${tan}\theta=\:\frac{\alpha}{{n}}\:\Rightarrow\:\theta={arctan}\left(\frac{\alpha}{{n}}\right) \\ $$$${we}\:{have} \\ $$$${A}_{{n}} ^{{n}} \:\:\:=\:\left(\mathrm{1}+\frac{\alpha^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)^{\frac{{n}}{\mathrm{2}}} \:\:\:.\begin{pmatrix}{{cos}\left({n}\theta\right)\:\:\:\:\:\:\:{sin}\left({n}\theta\right)}\\{−{sin}\left({n}\theta\right)\:\:\:\:\:{cos}\left({n}\theta\right.}\end{pmatrix} \\ $$$${but}\:\:\left(\mathrm{1}+\frac{\alpha^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)^{\frac{{n}}{\mathrm{2}}} \:={e}^{\frac{{n}}{\mathrm{2}}{ln}\left(\mathrm{1}+\frac{\alpha^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\:} \:\:\sim\:{e}^{\frac{\alpha^{\mathrm{2}} }{\mathrm{2}{n}}} \:\rightarrow\mathrm{1}\left({n}\rightarrow+\infty\right) \\ $$$${n}\:{arctan}\left(\frac{\alpha}{{n}}\right)\:\:\rightarrow\alpha\:\:{so}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} ^{{n}} \: \\ $$$$=\:\begin{pmatrix}{{cos}\alpha\:\:\:\:\:\:\:\:\:{sin}\alpha}\\{−{sin}\alpha\:\:\:\:\:\:{cos}\alpha}\end{pmatrix} \\ $$

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