Suppose first that $K$ is a finite extension of some $\mathbb Q_p$, with abs.
Galois gp. $G_K$.
A $p$-adic rep. of $G_K$ coming from geometry satisfies some basic conditions: it is pot. semi-stable, and the associated Weil--Deligne rep'n satisfies the Weil conjectures.
There are no other obvious conditions, and my memory (from a talk I saw many years ago, but perhaps it's written somewhere as well) is that Fontaine conjectured that these necessary conditions should be sufficient for an irred. $p$-adic rep. of $G_K$ to come from geometry. In the case when the rep'n looks like it should actually come from an abelian variety, I believe you can use Honda--Tate theory (and maybe some further related tools) to prove the conjecture, which gives some confidence in
the general case.
In the global case, one again has the obvious necessary conditions: finitely many ramified primes, pot. semi-stable locally at primes above $p$, and the Weil conjectures.
Again, there were no other obvious necessary conditions, and so, building on one's confidence in the local conjecture, it is natural to guess that they are also sufficient in the global context.
When Fontaine and Mazur were discussing this (the early 90's, I guess) Mazur pointed out that the condition on the Weil conjectures was not preserved under deformations, and so was an obstruction to ever proving such results. Mazur knew how to compute the expected dimensions of unrestricted local and global deformation rings, and Fontaine knew (at least in some case, such as the Fontaine--Laffaille case) how to compute the dimensions of local pot. semi-stable deformations rings.
If you imagine that the image of the global def. ring in the local rings meets the
pot. semi-stable locus transversely, then you find that the def. space of global reps. that are pot. semi-stable (of some given type and HT weights) is finite,
which fits with e.g. the Langlands reciprocity conjecture. (There are only finitely many Hecke eigenforms of fixed weight and level.)
These sorts of computations (I think there might be one in the original FM article) suggest that the conjecture might be true even without assuming the Galois rep.
satisfies the Weil conjectures, and give more confidence that it is true. To my mind, this deformation theoretic intuition gives pretty non-trivial motivation.
I'm not quite sure exactly how this origin story interacts with Wiles's proof.
I'm pretty sure that F and M made their conj. before Wiles's argument appeared,
and that deformation theory ideas were part of their motivation yoga. On the other
hand, clearly the whole conjecture became a lot more credible after Wiles's results.
As you note, much more is known about the conjecture now than was known when F and M first made their conjecture. This obviously adds to our confidence in it.
$\newcommand{\Z}{{\mathbb Z}}\newcommand{\Q}{{\mathbb Q}}
\newcommand{\N}{{\mathbb N}}$
Since $\hat\Z$ is metrizable,
it is sufficient to discuss the existence of the limit $\lim_{k\to\infty}u^{m_k}$
for a given invertible matrix $u\in M_d(\Q_\ell)$ and any sequence $m_k$, $k\in\N$ of integers converging to some element
$\hat n$ in $\hat\Z\setminus \Z$.
Consider a finite extension $L$ of $\Q_\ell$ containing all eigenvalues of $u$. It is well known that there exists a unique extension
of the $\ell$-adic valuation to $L$ which we call $v$. We will also use the integral domain $A\subset L$
of all elements of non-negative valuation, its maximal ideal $M$ of all elements of positive valuation and the
residue class field $F=A/M$. $F$ is a finite extension of the residue class field $\Z/\ell\Z$ of $\Z_\ell$. Hence $F$
is a finite field of $\ell^f$ elements with some $f\in\N$. Therefore $x^{\ell^f-1}=1$ for all $x\in F^*$. Hence $\mu^{\ell^f-1}\equiv 1\mod M$ for all $\mu\in A$ of valuation 0.
There is an invertible matrix $T\in M_d(L)$ such that $J=T^{-1}uT$ is in block diagonal form
$J=\mbox{bl.diag.}(J_1,\dots,J_r)$, where $J_i=\mu_i(I_{d_i}+Z_i)$ and $\mu_i$ is an eigenvalue of $u$
(hence non-zero), $I_{d_i}$ is the identity matrix of size $d_i$ and $Z_i$ is a nilpotent matrix, $Z_i^{d_i}=0$.
It is sufficient to study the convergence of $J_i^{m_k}$, $k\to\infty$, $i=1,\dots,r$ in $M_{d_i}(L)$.
We write $\hat n=(n_i+i\Z)_{i\in\N}$ where $n_i\equiv n_j\mod j$ if $j$ divides $i$.
$m_k\to\hat n$ means that for any $i\in\N$ there is a $K\in\N$ such that $m_k\equiv n_i\mod i$
for all $k\geq K$. In particular, $m_k$ converges in $\Z_l$ to $\check n$ determined by
$\check n\equiv n_{\ell^r}\mod \ell^r$ for all $r$. As a consequence, all binomial coefficients
$\binom{m_k}{s}$ converge in $L$ to $\binom{\check n}s$ as $k\to\infty$.
A first consequence of this for nilpotent $Z$ with $Z^d=1$ is that
$(I+Z)^{m_k}=I+\sum_{s=1}^{d-1}\binom{m_k}sZ^s$ converges
to $I+\sum_{s=1}^{d-1}\binom{\check n}sZ^s=:(I+Z)^{\check n}$.
It remains to study the convergence of $\mu^{m_k}$ for $\mu\in L$.
Case 1: $v(\mu)\neq0$: Then clearly $\mu^{m_k}$ cannot converge as
$m_k$ and hence $v(\mu^{m_k})$ does not remain bounded.
Case 2: $v(\mu)=0$: Since $\mu^Q\equiv 1\mod M$ for $Q=\ell^f-1$, it is convient to write
$m_k=n_Q+Q\tilde m_k$ for all sufficiently large $k$. This is possible because
$m_k\equiv n_Q\mod Q$ for large $k$.
Clearly, $\tilde m_k$ converges to some limit $\hat{\tilde n}$ in $\hat Z$ and hence also to
$\check{\tilde n}$, say, in $Z_\ell$.
We have $\mu^{m_k}=\mu^{n_Q}(1+d)^{\tilde m_k}$ where $d=\mu^{Q}-1$ has positive valuation.
The convergence of $(1+d)^{\tilde m_k}$ follows from the expansions
$$(1+d)^{\tilde m_k}=1+\sum_{s=1}^\infty \binom{\tilde m_k}s\,d^s\mbox{ and }
\binom{\tilde m_k}s\to\binom{{\check{\tilde n}}}s\mbox{ for all }s.$$
Combining the above statements shows that $u^{m_k}$ converges in $M_d(L)$ and hence in
$M_d(\Q_\ell)$ as $k\to\infty$ if and only if all eigenvalues of $u$ have valuation 0.
Best Answer
That's not what the proposition says; it says that the eigenvalues are units. They don't have to lie in $\mathbb{Q}_l$.
Here's a hint. If an invertible matrix $M$ acting on $V$ has an eigenvalue which is not a unit, then, after possibly inverting $M$, we may assume it has an eigenvalue $\lambda$ with $|\lambda| > 1$. But then $M^{n}$ for large $n$ will have an eigenvalue $\lambda^n$ where $|\lambda^n|$ is getting bigger and bigger.
Now you have to understand (given the structure of $\widehat{\mathbb{Z}}$) why continuity of $G_K \rightarrow \mathrm{GL}(V)$ implies that there will be some big integral $n$ such that $M^n$ and $M^0 = I$ are as close as you wish linear maps, and this will preclude the eigenvalues of $M^n$ being big.