~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 \( bc > 0 \) $$ \begin{bmatrix} \cosh(\sqrt{bc}) & \sqrt{b/c}\sinh(\sqrt{bc}) \\ \sqrt{c/b} \sinh(\sqrt{bc}) & \cosh(\sqrt{bc}) \end{bmatrix}.$$