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

$$letgive{A}_n=\left(\begin{array}{c}1\frac{\alpha }{n}\\ -\frac{\alpha }{n}1\end{array}\right)$$$$calculate{lim}_{n\to +\mathrm{\infty }}{A}_n^n.$$

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

$$wehave\sqrt{1+\frac{{\alpha }^2}{n^2}}=\frac{1}{n}\sqrt{n^2+{\alpha }^2}\Rightarrow$$$$A_n=\sqrt{1+\frac{{\alpha }^2}{n^2}}.\left(\begin{array}{c}\frac{1}{\sqrt{1+\frac{{\alpha }^2}{n^2}}}\frac{\alpha }{n\sqrt{1+\frac{{\alpha }^2}{n^2}}}\\ -\frac{\alpha }{n\sqrt{1+\frac{{\alpha }^2}{n^2}}}\frac{1}{\sqrt{1+\frac{{\alpha }^2}{n^2}}}\end{array}\right)$$$$A_n=\sqrt{1+\frac{{\alpha }^2}{n^2}}\left(\begin{array}{c}cos{\theta }_{}sin\theta \\ -sin\theta cos\theta \end{array}\right)$$$$withcos\theta =\frac{1}{\sqrt{1+\frac{{\alpha }^2}{n^n}}}andsin\theta =\frac{\alpha }{n.\sqrt{1+\frac{{\alpha }^2}{n^2}}}\Rightarrow$$$$tan\theta =\frac{\alpha }{n}\Rightarrow \theta =arctan\left(\frac{\alpha }{n}\right)$$$$wehave$$$$A_n^n={(1+\frac{{\alpha }^2}{n^2})}^{\frac{n}{2}}.\left(\begin{array}{c}cos\left(n\theta \right)sin\left(n\theta \right)\\ -sin\left(n\theta \right)cos(n\theta \end{array}\right)$$$$but{(1+\frac{{\alpha }^2}{n^2})}^{\frac{n}{2}}=e^{\frac{n}{2}ln(1+\frac{{\alpha }^2}{n^2})}\sim e^{\frac{{\alpha }^2}{2n}}\to 1(n\to +\mathrm{\infty })$$$$narctan\left(\frac{\alpha }{n}\right)\to \alpha so{lim}_{n\to +\mathrm{\infty }}{A}_n^n$$$$=\left(\begin{array}{c}cos\alpha sin\alpha \\ -sin\alpha cos\alpha \end{array}\right)$$

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