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Finding $\lim_{n\rightarrow \infty}\begin{pmatrix} 1 & \frac{x}{n}\\ \\ -\frac{x}{n} & 1 \end{pmatrix}^n$ for all $x\in \mathbb{R}$

Try:
Let $$ A = \begin{pmatrix}1&\frac{x}{n}\\\\-\frac{x}{n}&1\end{pmatrix}.$$

Then $$ A^2 = \begin{pmatrix}1-\frac{x^2}{n^2}&\frac{2x}{n}\\\\-\frac{2x}{n}&1-\frac{x^2}{n^2}\end{pmatrix}$$

And then $$A^3 = \begin{pmatrix}1-3\frac{x^2}{n^2}&\frac{3x}{n}-\frac{x^3}{n^3}\\\\-3\frac{x}{n}+\frac{x^3}{n^3}&1-\frac{x^2}{n^2}\end{pmatrix}$$

So by using same way and taking $\lim_{n\rightarrow \infty}A^n = \begin{pmatrix}1&0\\\\0&1\end{pmatrix}$

But answer given as $$\begin{pmatrix}\cos x &\sin x\\\\-\sin x&\cos x\end{pmatrix}$$

Could some help me where I am missing and also explain how to solve it?

DXT
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2 Answers2

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We have $$A = \sqrt{1 + \frac {x^2}{n^2}}\begin{pmatrix}\frac 1 {\sqrt{1 + \frac {x^2}{n^2}}}&\frac {\frac x n} {\sqrt{1 + \frac {x^2}{n^2}}}\\\\-\frac {\frac x n} {\sqrt{1 + \frac {x^2}{n^2}}}&\frac 1 {\sqrt{1 + \frac {x^2}{n^2}}}\end{pmatrix}$$ So if $\theta_n$ is such that $\cos \theta_n=\frac 1 {\sqrt{1 + \frac {x^2}{n^2}}}$ and $\sin \theta_n = -\frac {\frac x n} {\sqrt{1 + \frac {x^2}{n^2}}}$, and $\rho_n=\sqrt{1 + \frac {x^2}{n^2}}$, then $$A=\rho_n\begin{pmatrix}\cos \theta_n&-\sin \theta_n\\\\\sin \theta_n&\cos \theta_n\end{pmatrix}$$ Therefore, $A$ is the product of the scaling factor $\rho_n$ with the rotation matrix of angle $\theta_n$. So powers of $A$ are $$A^n=\rho_n^n\begin{pmatrix}\cos n\theta_n&-\sin n\theta_n\\\\\sin n\theta_n&\cos n\theta_n\end{pmatrix}$$ Now, using Taylor expansions, $$\lim_{n\rightarrow +\infty}\rho_n^n=1$$ and $$\cos(n\theta_n)=\cos\left(n\arccos\bigg(\frac 1 {\sqrt{1 + \frac {x^2}{n^2}}}\bigg)\right)\rightarrow\cos(x)$$ $$\sin(n\theta_n)=\sin\left(n\arcsin\bigg(\frac {-\frac {x} n} {\sqrt{1 + \frac {x^2}{n^2}}}\bigg)\right)\rightarrow-\sin(x)$$

Stefan Lafon
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You seem to be arguing that the terms proportional to $x$, $x^2$, $x^3$, etc. vanish in the limit $n \to \infty$. But the problem with this argument is that the numerical coefficients in front of these terms also grow as $n \to \infty$, and diverge without bound. To see this, define $$ A_n = \begin{bmatrix}1 & x/n\\ -x/n & 1 \end{bmatrix}^n. $$ Here are the terms in the upper-left entry for the first few values of $n$: \begin{align} (A_1)_{12} &= \frac{x}{n} = x \\ (A_2)_{12} &= \frac{2x}{n} = x\\ (A_3)_{12} &= \frac{3x}{n} - \frac{x^3}{n^3} = x - \frac{x^3}{27} \\ (A_4)_{12} &= \frac{4 x}{n} - \frac{4 x^3}{n^3} = x - \frac{x^3}{16} \\ (A_5)_{12} &= \frac{5 x}{n} - \frac{10 x^3}{n^3} + \frac{x^5}{n^5} = x - \frac{10 x^3}{125} + \frac{x^5}{3125}\\ \end{align} Just looking at the term proportional to $x$ in this sequence, it seems pretty likely that this will not vanish as $n \to \infty$.

Michael Seifert
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