Reaching the speed of light via quantum mechanical uncertainty? Suppose you accelerate a body to very near the speed of light $c$ where $v = c - \epsilon$.  Although this would take an enormous energy, is it possible the last arbitrarily small velocity needed -- $\epsilon$ -- could be overcome with a minor bump in velocity due to the uncertainty principle?
 A: No, because the uncertainty principle operates between position and momentum rather than position and velocity. For speeds much less than $c$, momentum is just proportional to velocity: $p = mv$. But at relativistic speeds we have to use the relativistic version,
$$
p = \gamma mv,
$$
where $\gamma = 1/\sqrt{1-v^2/c^2}$. Substituting this in and squaring both sides we get
$$
p^2 = \frac{m^2v^2}{1-{v^2}/{c^2}},
$$
which we can rearrange a little to get
$$
v^2 = \frac{p^2}{ m^2 + p^2/c^2 },
$$
or
$$
v = \frac{p}{\sqrt{ m^2 + p^2/c^2 }}.
$$
Now, the limit of this as $p \to \infty$ is just
$$
v = \frac{p}{\sqrt{p^2/c^2 }} = c.
$$
The momentum $p$ can fluctuate due to the uncertainty principle, but now you can see that now matter how big $p$ gets, $v$ will always be less than $c$.
A: No. First of all, Planck's constant is not a speed, so you can't compute $c - \hbar$. But you can reword the question to get around that problem, something like this:

Is there some speed $\epsilon$ such that an object traveling at speed $c - \epsilon$ could experience a quantum fluctuation that temporarily takes its speed to greater than $c$?

The answer to this is still no. Now, in order to really understand why, you could dig into the details of quantum field theory, and learn the meaning of the statement "local operators separated by spacelike intervals commute" which is, in some sense, the most fundamental reason. But I'm guessing that'd be more detail than you're looking for.
As a simplified (but still basically accurate) explanation, you can use the same argument for why you can't bump a classical object moving at speed $c - \epsilon$ up to speed greater than $c$ by giving it a little push. That reason is that when something speeds up, spacetime "rotates" around it, but in such a way that all trajectories with speeds less than $c$ continue to have speeds less than $c$. In particular, this rotation (the Lorentz boost) transforms a trajectory with speed $v$ into a trajectory with speed $\frac{v + \Delta v}{1 + v\Delta v/c^2}$. No matter how close you are to the speed of light, speeding up will only take you a fraction of the way closer to $c$, and that is just as true for a quantum fluctuation as it is for a classical push.
A: Quantum uncertainty as shown by Heisenberg precludes the information from reaching us faster than the speed c for light. 
Lorentz invariance is preserved by the existence of antimatter. This was shown by Dirac. Thus as ( Feynman? ) some noticed, you can describe a positron as an electron moving backward in time. 
So the speed of light is not classical so while the speed at which information can move in space is certain the lack of certainty of velocity and position in QM and the ability to see it differently as a particle or antiparticle depending on your perspective keeps everything consistent. If you are looking for more rigor I leave it to others but this works for me to avoid paradoxes
