For example
EDIT (Atending OP´s objection) The important thing is that as stated, consider your favourite matrix which does not verify the statement, and consider problem is reduced to a map linear algebra problem since it is possible to construct a diffeomorphism of $\mathbb{R}^n$ such that matrix $J_n(0)$ for the space which orbit of $0$ is the identity outside product of any sequence of invertible matrices.
To construct this, consider a ball translation of $B$ \mathbb{R}^n$ (say $F(x)= x+b$) and this linear map in a neighborhood of zero. Now compose this map with a translation of big enough modulus $nb$ modify the diffeomorphism so that the image derivative in that point is the desired matrix $A_n$.
In dimension $2$, a way of getting the ball desired counterexample is disjoint from itselfto consider the two times two upper triangular matrices $A_n$ with both eigenvalues $1$ and $K^n$ in the upper right corner (I am not being able to write matrices).
The iterates
We get that the eigenvalues of $0$ J_n$ will verify that the Jacobian matrix is be always the same (since after one iteration, one starts to multiply by $1$, but the identity)norm of $J_n$ grows exponentially, so, it will never verify your statementis not true that the limits coincide.
I haven't thought on how to make a counterexample where the norms of $A_n$ are bounded but it should be not very difficult.
However, when some recurrence is added into the game, some results in the direction of what you are looking for are available. A key word for searching is Oseledets Theorem (or Multiplicative ergodic theorem, notice that in some places it is named Oseledec, or with some variations, othe key word for searching is: Lyapunov exponents). In particular, given an invariant probability measure, what you look for is satisfied for almost every point.
Playing with the proofs of this results, other more ``natural'' counterexamples can be constructed.
