Given a Matrix X and another Matrix $M=e^X$ and $\det M =1$, I want to show that the Trace of X vanishes, e. g. $\mathrm{Tr} (X) =0$.
I think that one can write it as following:
\begin{equation} \begin{split} \det M & = \det e^X \\ & = \det [\lim_{n \to \infty}(1+X/n)^n] \\ & = \lim_{n \to \infty}(\det ((1+X/n)^n) \\ & = \lim_{n \to \infty}(1+ \mathrm{Tr} (X))^n, \end{split} \end{equation} where I used $\det(X^n)=(\det(X))^n$ in the last step. From the claim $\det M =1$, $\mathrm{Tr} (X)$ needs to be zero.
I am not sure about changing the order of determinant and limit. Why can one do that?
If $M$ is an $n\times n$ matrix, then you can consider the determinant to be a homogeneous polynomial function in the entries of the matrix. So if you take the limit of the determinant, you're basically taking the limit of a function of $n$ that is determined by its coordinates. So provided you have a sequence of square matrices $M(n)$ in which each entry is a function of $n$ that converges, entrywise, to a matrix $M$, then the determinant and limits should commute.
As a simple example, consider $$M(n)=\begin{pmatrix} a(n) & b(n) \\ c(n) & d(n) \end{pmatrix} \to_{n\to\infty} M=\begin{pmatrix} A & B \\ C & D \end{pmatrix},$$ and note that $$\lim_n \det(M(n)) = \lim_n a(n)d(n) - b(n)c(n) = AD - BC = \det(M) = \det(\lim_n M_n).$$ For higher dimensions, you just get a more complicated formula for $\det(M(n))$, since there will be many more terms.
As a side note, you could also write $$e^X = \lim_{n\to\infty} \sum_{k=0}^n \frac{X^n}{n!}$$ (this is how I've seen it defined, usually).