# ~blin's tilde

## Can you solve this?

$$\lim_{n \to \infty} \begin{bmatrix} \cos(a/n) & \sin(b/n) \\ \sin(c/n) & \cos(d/n) \end{bmatrix}^n = \text{?}$$
My solution Observe that $$\begin{bmatrix} \cos(a/n) & \sin(b/n) \\ \sin(c/n) & \cos(d/n) \end{bmatrix}^n = \left( E + \frac{1}{n} \begin{bmatrix} n(\cos(a/n)-1) & n\sin(b/n) \\ n\sin(c/n) & n(\cos(d/n)-1) \end{bmatrix}\right)^n.$$ Now let $$A_n := \begin{bmatrix} n(\cos(a/n)-1) & n\sin(b/n) \\ n\sin(c/n) & n(\cos(d/n)-1) \end{bmatrix}.$$ Then $$\lim_{n \to \infty} A_n = \begin{bmatrix} 0 & b \\ c & 0 \end{bmatrix} =: A.$$ For large enough $$n$$, we may apply the matrix logarithm to the equality above and expand it into a series to obtain $$\lim_{n \to \infty} n \ln\left(E + \frac{1}{n} A_n\right) = \lim_{n \to \infty} \left( A_n + \frac{1}{n} \sum_{k = 2}^{\infty} \frac{(-1)^{k+1}}{k} \left( \frac{A_n^k}{n^{k-1}} \right) \right) = A.$$ By continuity of the matrix exponential, the limit we are looking for is hence the exponential of the matrix $$A$$, the value of which depends on its eigenvalues, which are the roots of the polynomial $$x^2 - bc = 0$$. We consider only the case $$bc > 0$$, the others are similar. In this case, the eigenvalue decomposition of $$A$$ is calculated to be $$A = \begin{bmatrix} \sqrt{b} & \sqrt{b} \\ \sqrt{c} & -\sqrt{c} \end{bmatrix}\begin{bmatrix} \sqrt{bc} & 0 \\ 0 & \sqrt{bc} \end{bmatrix} \frac{1}{2}\begin{bmatrix} 1/\sqrt{b} & 1/\sqrt{c} \\ 1/\sqrt{b} & -1/\sqrt{c} \end{bmatrix}.$$ Using that conjugation of matrices commutes with the exponential, we obtain $$\exp(A) = \frac{1}{2} \begin{bmatrix} e^{\sqrt{bc}} + e^{-\sqrt{bc}} & \sqrt{b/c} (e^{\sqrt{bc}} - e^{-\sqrt{bc}}) \\ \sqrt{c/b}(e^{\sqrt{bc}} - e^{-\sqrt{bc}}) & e^{\sqrt{bc}} + e^{-\sqrt{bc}} \end{bmatrix}$$ and hence our final solution in the case $$bc > 0$$ $$\begin{bmatrix} \cosh(\sqrt{bc}) & \sqrt{b/c}\sinh(\sqrt{bc}) \\ \sqrt{c/b} \sinh(\sqrt{bc}) & \cosh(\sqrt{bc}) \end{bmatrix}.$$