Doesn't the speed of light limit imply the same electron can be annihilated twice? If I understand correctly, there is a small probability the same electron to be found anywhere in the universe.
Suppose that an anti-electron collides with an electron, annihilating it and producing two photons. Assuming the speed of light limit is correct, right after the collision, the probability that the electron is found at a distance of $d$ from the collision must still be non-zero for a time of at least $d/c$.
But wouldn't this mean that it's still possible for the annihilated electron to still be present somewhere else and collide with another anti-electron, meaning that the same electron was annihilated twice?
 A: What you need to realize, is that quantum mechanics throws the concept of local realism out of the window. When you measure one entangled particle, you instantly change the state of the other.
With your electron, annihilation with a localized positron is a measurement of locality. And the result of that measurement instantly changes the entire wave function of the electron (I'm using the Copenhagen Interpretation here). The other possible annihilation is another measurement of locality on the same wave function, and if either annihilation succeeds, the other won't.
Note that this does not mean that you can communicate any information faster than light: You cannot tell from the results of the measurements which measurement happened first on an entangled state, you can only tell that the two measurements influence each other.
If you are uneasy about this "spooky action at a distance", congrats, you are in good company. Einstein had similar concerns. He and others came up with the EPR paradox to show that quantum mechanics could not be complete because it predicted such "spooky action at a distance", which simply could not be. Unfortunately for these three brilliant physicists, later experiments have shown that the predictions of quantum mechanics are indeed correct, and that no theory with local realism can adequately describe reality.
A: The nice answer by knzhou considers a situation in which the electron is annihilated with probability $1$. The following answer considers a situation in which the final state is a superposition of already-annihilated and not-yet-annihilated.
Choose two points in space, $x_1$ and $x_2$, that are arbitrarily far away from each other. Consider an electron in the state
$$
 |\psi\rangle = |1\rangle + |2\rangle,
$$
where


*

*$|1\rangle$ is a state in which the electron is tightly localized near $x_1$, and an antielectron is approaching $x_1$.

*$|2\rangle$ is a state in which the electron is tightly localized near $x_2$, and an antielectron is approaching $x_1$. (This is not a typo! The antielectron is approaching $x_1$ in both $|1\rangle$ and $|2\rangle$.)
When the antielectron reaches the point $x_1$, 


*

*$|1\rangle$ becomes $|1'\rangle$, a state with two photons propagating outward from $x_1$.

*$|2\rangle$ becomes $|2'\rangle$, a state with one electron at $x_2$ and one antielectron receding from $x_1$.
Time-evolution is linear, so the final state overall is
$$
 |\psi'\rangle = |1'\rangle + |2'\rangle.
$$
This is a superposition of "electron already annihilated" and "electron still present at $x_2$," so there is still a non-zero probability 
$$
 \frac{\langle 2'|2'\rangle}{\langle\psi'|\psi'\rangle}
$$
that an antielectron can annihilate the electron at $x_2$. 
Be careful, though: this does not mean that the same electron can be annihilated twice! The electron is only annihilated once, but it can be in a state that is a superposition of already-annihilated and not-yet-annihilated. If we measure some observable to tell us how many times the electron was annihilated (say, by counting pairs of outgoing photons), the answer will be either $0$ or $1$, with probabilities calculated as shown above.
A: No, you have to apply the superposition principle consistently. Schematically, the initial state is
$$|\text{electron here} \rangle + |\text{electron there} \rangle$$
where I've dropped normalization constants, and a $+$ denotes quantum superposition. Now suppose a lot of positrons come through, so electron states get annihilated, 
$$|\text{electron here} \rangle \mapsto |\text{some gamma rays here} \rangle,$$
$$|\text{electron there} \rangle \mapsto |\text{some gamma rays there} \rangle.$$
What you're essentially claiming is that the final state is 
$$|\text{some gamma rays here } \textbf{and} \text{ some gamma rays there}\rangle$$
but if you just apply linearity, the final state is actually
$$|\text{some gamma rays here} \rangle + |\text{some gamma rays there} \rangle.$$
This reasoning implies you can't get double the gamma rays, no matter how severe the speed of light delay or any other delays are. Instead the superposition of electron positions can at best turn into a superposition of gamma ray positions. The mistake you made is essentially forgetting that the electromagnetic field behaves quantum mechanically too. (It's a forgivable mistake, which was made by many in the early days of quantum mechanics, leading to precisely the same kinds of paradoxes.)
