The continuum limit of the path integral & differential operators When deriving the path integral formulation at one reaches an expression of the form
$$
\intop \prod_n dx_n \prod_n dp_n  e^{\frac{i}{\hbar } \Delta t \sum_{n} (p_{n+1} \frac{x_{n+1}-x_{n}}{\Delta t}-T(p_n) -\frac{V(x_{n+1})+V(x_n)}{2})} 
$$
Where $V(x)$ ar the matrix elements of the potential operator $V(\hat{X})$, and similary for $T(\hat{P})$ the kinetic energy operator. Many texts simply state that at the continuum limit this becomes
$$
\intop D x D p  \, e^{\frac{i}{\hbar} \intop dt(p\dot{x}- H(p,x)}. 
$$
What is not clear to me is the justification to take the limit $$\frac{x_{n+1}-x_{n}}{\Delta t} \rightarrow \dot{x}$$ since $x_{n+1}$ and $x_{n}$ are two independent integration variables and there is no obligation that they approach each other when the limit is taken.
Some sources (for example Altland and Simons p.99) mention briefly that paths where  $x_{n+1} - x_{n} \not\rightarrow 0$ cancel each other due to the rapid phases approximation. Other simply avoid the issue ( or Kleinert ch. 2, Feznman and Hibbs ch. 2). Is there a way to see it clearly? What assumption suffice to validly take the limit?
 A: 
What is not clear to me is the justification to take the limit $$\frac{x_{n+1}-x_{n}}{\Delta t} \rightarrow \dot{x}$$ since $x_{n+1}$ and $x_{n}$ are two independent integration variables and there is no obligation that they approach each other when the limit is taken.

If they don't, then the velocity is infinite. For finite spacing, the terms where $x_{n+1}$ is very far away from $x_n$ have very high "velocity."

Some sources (for example Altland and Simons p.99) mention briefly that paths where  $x_{n+1} - x_{n} \not\rightarrow 0$ cancel each other due to the rapid phases approximation. Other simply avoid the issue ( or Kleinert ch. 2, Feznman and Hibbs ch. 2). Is there a way to see it clearly?

When the velocity is infinite (or very high), you can see or expect that the $e^{ip\dot x}$ term has a "rapidly changing" phase.
A: There can be no good justification for $\frac{x_{n-1} - x_n}{\Delta t}\to \dot{x}$ because that's not what happens in the actual mathematically rigorous path integral! You can try to do various physical handwaves for this, but in the end the truth is that the derivative notation here used by physicists is just a shorthand, not something that has an actual formal meaning.
The proper mathematical definition of the integration in the quantum mechanical path integral is by defining the Wiener measure as representing the path integral of the kinetic part, i.e. there is a measure $\mathrm{d}W_{q,q'}$ induced by the free Hamiltonian $H \propto p^2$ on all continuous paths with endpoints $q,q'$ and one indeed defines this for finite $n$ by a very similar formula but one never needs to take an actual limit $n\to \infty$ where the difference quotients become derivatives. Instead the argument that this defines measure everywhere involves standard mathematical arguments about the uniqueness of Gaußian measures, which "in spirit" correspond to this limit but cruically do not do so formally.
The Wiener measure works like this: The kernel of the free Hamiltonian is
$$K_t(q,q') = (2\pi t)^{3/2}\mathrm{e}^{-(q -q')^2/2t}$$
meaning a solution $\psi(x,t)$ to the free Schrödinger equation fulfills
$$ \psi(q,t) = \int K_t(q,q')\psi(q',0)\mathrm{d}q'$$
and we define the measure of a set $S_{I,t_1}$ of continuous paths $q(t)$ with $t\in[0,t_0]$ and with $q(t_1)\in I\subset \mathbb{R}^3$ for $I$ a nice (Borel) set to be
$$ \int_{S_{I,t_1}}\mathrm{d}W_{q,q'} := \int_I K_{t_0/2 + t_1}(q,q_1)K_{t_0/2-t_1}(q_1,q')\mathrm{d}q_1.$$
Physicists write $\mathrm{d}W_{q,q'} = \mathcal{D}q\mathrm{e}^{-S_0[q(t)]}$ where $S_0$ is the action of a free particle. For an interacting action with potential $V(q)$, we simply integrate $\mathrm{e}^{-V(q(t))}\mathrm{d}W_{q,q'}$ - as long as  the potential does not involve derivatives, this works just fine without ever talking about differentiability.
Note that this definition does not involve anywhere the actual paths $q(t)$ - it's just a specific convolution of the function $K_t(q,q')$. This might strike you as strange, but there is nothing mathematically wrong with this definition. What this "means" is that we're saying that the paths that pass through $I$ at an instant $t_1$ are weighted by the probability first for them to evolve from $q(0) = q$ and then by the probability to then evolve from some point $q_1 = q(t_1)\in I$ to the fixed endpoint $q(t_0) = q'$. The integral is just the convolution of these two probability densities.
Crucially, the Wiener measure is a measure on all continuous paths and not on differentiable paths (in fact, the differentiable paths have measure zero), and so the object $\dot{x}$ doesn't actually make any sense as something that appears in the integrand - a merely continuous path doesn't have a $\dot{x}$ everywhere.
While the proper mathematical construction of the Wiener measure and relating it to quantum mechanical amplitudes (the Feynman-Kac formula) does involve a limit that takes $n\to\infty$, it never needs to assume differentiability of the path - there is only an argument about Riemann integrability of the potential $V[q(t)]$, which is much weaker.
For the actual details on how to construct the path integral rigorously instead of the vague outline above, I recommend chapter 3 of "Quantum Physics" by Glimm and Jaffe.
