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The right formula is $$\Delta X \Delta P \geq h/4\pi$$ where $P$ is the momentum which is approximatively $mv$ only for small velocities $v$ when compared with $c$. Otherwise you have to use the relativistic expression $$P = mv/ \sqrt{1-v^2/c^2}.$$ If $\Delta X$ is small, then $\Delta P$ is large but, according to the formula above, the velocity remains of the order of $c$ at most. That is because, in the formula above, $P\to +\infty$ corresponds to $v\to c$.

With some details, solving the above identity for $v$, we have $$v = \frac{P}{m \sqrt{1+ P^2/m^2c^2}}\:,$$ so that $$v\pm \Delta v = \frac{P\pm \Delta P}{m \sqrt{1+ (P\pm \Delta P)^2/m^2c^2}}$$ where $$\Delta P = \frac{\hbar}{2\Delta X}\:.$$ For a fixed value of $P$ and $\Delta X \to 0$, we have $$v\pm \Delta v = \lim_{\Delta P \to + \infty}\frac{P\pm \Delta P}{m \sqrt{1+ (P \pm \Delta P)^2/m^2c^2}}= \pm c\:.\tag{1}$$

Finally, it is not difficult to see that (using the graph of the hyperbolic tangent function) $$-1 \leq \frac{(P\pm \Delta P)/mc}{ \sqrt{1+ (P \pm \Delta P)^2/m^2c^2}}\leq 1\tag{2}\:.$$ We therefore conclude that $$-c \leq v\pm \Delta v \leq c,$$ where the boundary values are achieved only for $\Delta X \to 0$ according to (1). Relativity is safe...

The right formula is $$\Delta X \Delta P \geq h/4\pi$$ where $P$ is the momentum which is approximatively $mv$ only for small velocities $v$ when compared with $c$. Otherwise you have to use the relativistic expression $$P = mv/ \sqrt{1-v^2/c^2}.$$ If $\Delta X$ is small, then $\Delta P$ is large but, according to the formula above, the speed remains of the order of $c$ at most. That is because, in the formula above, $P\to +\infty$ corresponds to $v\to c$.

With some details, solving the above identity for $v$, we have $$v = \frac{P}{m \sqrt{1+ P^2/m^2c^2}}\:,$$ so that $$v\pm \Delta v = \frac{P\pm \Delta P}{m \sqrt{1+ (P\pm \Delta P)^2/m^2c^2}}.$$ We have obtained the exact expression of $\Delta v$: $$\pm \Delta v = \frac{P\pm \Delta P}{m \sqrt{1+ (P\pm \Delta P)^2/m^2c^2}} - \frac{P}{m \sqrt{1+ P^2/m^2c^2}},$$ where $$\Delta P = \frac{\hbar}{2\Delta X}\:.$$ This is a complicated expression but it is easy to see that the final speed cannot exceed $c$ in any cases. For a fixed value of $P$ and $\Delta X \to 0$, we have $$v\pm \Delta v = \lim_{\Delta P \to + \infty}\frac{P\pm \Delta P}{m \sqrt{1+ (P \pm \Delta P)^2/m^2c^2}}= \pm c\:.\tag{1}$$

Finally, it is not difficult to see that (using the graph of the hyperbolic tangent function) $$-1 \leq \frac{(P\pm \Delta P)/mc}{ \sqrt{1+ (P \pm \Delta P)^2/m^2c^2}}\leq 1\tag{2}\:.$$ We therefore conclude that $$-c \leq v\pm \Delta v \leq c,$$ where the boundary values are achieved only for $\Delta X \to 0$ according to (1). Relativity is safe...

The right formula is $$\Delta X \Delta P \geq h/4\pi$$ where $P$ is the momentum which is approximatively $mv$ only for small velocities $v$ when compared with $c$. Otherwise you have to use the relativistic expression $$P = mv/ \sqrt{1-v^2/c^2}.$$ If $\Delta X$ is small, then $\Delta P$ is large but, according to the formula above, it remains of the order of $c$ at most. That is because, in the formula above, $P\to +\infty$ corresponds to $v\to c$.

With some details, solving the above identity for $v$, we have $$v = \frac{P}{m \sqrt{1+ P^2/m^2c^2}}\:,$$ so that $$v\pm \Delta v = \frac{P\pm \Delta P}{m \sqrt{1+ (P\pm \Delta P)^2/m^2c^2}}$$ where $$\Delta P = \frac{\hbar}{2\Delta X}$$ For a fixed value of $P$ and $\Delta X \to 0$, we have $$v\pm \Delta v = \lim_{\Delta P \to + \infty}\frac{P\pm \Delta P}{m \sqrt{1+ (P \pm \Delta P)^2/m^2c^2}}= \pm c\:.\tag{1}$$

Finally, it is not difficult to see that (using the graph of the hyperbolic tangent function) $$-1 \leq \frac{(P\pm \Delta P)/mc}{ \sqrt{1+ (P \pm \Delta P)^2/m^2c^2}}\leq 1\tag{2}\:.$$ We therefore conclude that $$-c \leq v\pm \Delta v \leq c,$$ where the boundary values are achieved only for $\Delta X \to 0$ according to (1).

The right formula is $$\Delta X \Delta P \geq h/4\pi$$ where $P$ is the momentum which is approximatively $mv$ only for small velocities $v$ when compared with $c$. Otherwise you have to use the relativistic expression $$P = mv/ \sqrt{1-v^2/c^2}.$$ If $\Delta X$ is small, then $\Delta P$ is large but, according to the formula above, the velocity remains of the order of $c$ at most. That is because, in the formula above, $P\to +\infty$ corresponds to $v\to c$.

With some details, solving the above identity for $v$, we have $$v = \frac{P}{m \sqrt{1+ P^2/m^2c^2}}\:,$$ so that $$v\pm \Delta v = \frac{P\pm \Delta P}{m \sqrt{1+ (P\pm \Delta P)^2/m^2c^2}}$$ where $$\Delta P = \frac{\hbar}{2\Delta X}\:.$$ For a fixed value of $P$ and $\Delta X \to 0$, we have $$v\pm \Delta v = \lim_{\Delta P \to + \infty}\frac{P\pm \Delta P}{m \sqrt{1+ (P \pm \Delta P)^2/m^2c^2}}= \pm c\:.\tag{1}$$

Finally, it is not difficult to see that (using the graph of the hyperbolic tangent function) $$-1 \leq \frac{(P\pm \Delta P)/mc}{ \sqrt{1+ (P \pm \Delta P)^2/m^2c^2}}\leq 1\tag{2}\:.$$ We therefore conclude that $$-c \leq v\pm \Delta v \leq c,$$ where the boundary values are achieved only for $\Delta X \to 0$ according to (1). Relativity is safe...

The right formula is $$\Delta X \Delta P \geq h/4\pi$$ where $P$ is the momentum which is approximatively $mv$ only for small velocities $v$ when compared with $c$. Otherwise you have to use the relativistic expression $$P = mv/ \sqrt{1-v^2/c^2}.$$ If $\Delta X$ is small, then $\Delta P$ is large but, according to the formula above, it remains of the order of $c$ at most. That is because, in the formula above, $P\to +\infty$ corresponds to $v\to c$.

With some details, solving the above identity for $v$, we have $$v = \frac{P}{m \sqrt{1+ P^2/m^2c^2}}\:,$$ so that $$v\pm \Delta v = \frac{P\pm \Delta P}{m \sqrt{1+ (P+ \Delta P)^2/m^2c^2}}$$ where $$\Delta P = \frac{\hbar}{2\Delta X}$$ For a fixed value of $P$ and $\Delta X \to 0$, we have $$v\pm \Delta v = \lim_{\Delta P \to + \infty}\frac{P\pm \Delta P}{m \sqrt{1+ (P \pm \Delta P)^2/m^2c^2}}= \pm c\:.$$

The right formula is $$\Delta X \Delta P \geq h/4\pi$$ where $P$ is the momentum which is approximatively $mv$ only for small velocities $v$ when compared with $c$. Otherwise you have to use the relativistic expression $$P = mv/ \sqrt{1-v^2/c^2}.$$ If $\Delta X$ is small, then $\Delta P$ is large but, according to the formula above, it remains of the order of $c$ at most. That is because, in the formula above, $P\to +\infty$ corresponds to $v\to c$.

With some details, solving the above identity for $v$, we have $$v = \frac{P}{m \sqrt{1+ P^2/m^2c^2}}\:,$$ so that $$v\pm \Delta v = \frac{P\pm \Delta P}{m \sqrt{1+ (P\pm \Delta P)^2/m^2c^2}}$$ where $$\Delta P = \frac{\hbar}{2\Delta X}$$ For a fixed value of $P$ and $\Delta X \to 0$, we have $$v\pm \Delta v = \lim_{\Delta P \to + \infty}\frac{P\pm \Delta P}{m \sqrt{1+ (P \pm \Delta P)^2/m^2c^2}}= \pm c\:.\tag{1}$$

Finally, it is not difficult to see that (using the graph of the hyperbolic tangent function) $$-1 \leq \frac{(P\pm \Delta P)/mc}{ \sqrt{1+ (P \pm \Delta P)^2/m^2c^2}}\leq 1\tag{2}\:.$$ We therefore conclude that $$-c \leq v\pm \Delta v \leq c,$$ where the boundary values are achieved only for $\Delta X \to 0$ according to (1).