Why isn't there a limit for a Euclidean rotation, as for a Minkowski rotation?

From invariance of the Minkowski scalar product, we get the Lorentz transformations. In addition, we get a constant $c$ preventing space-like and time-like intervals being rotated into one another.

The Euclidean transformations are derived in the same way from invariance of the Euclidean scalar product, yet there isn't a constant that limits these rotations as above. Why?

The natural way to parametrize a boost in spacetime (hyperbolic geometry) is by the quantity known as rapidity, just as the natural way to parametrize a rotation in Euclidean space is the angle.

An angle measures distance along a unit circle. That is, the circumference of a unit circle is $2\pi$, so we parametrize angles by the distance the angle subtends on the unit circle.

Rapidity measures invariant interval along the unit hyperbola. It is the natural way of parametrizing distance along the unit hyperbola, and boosts are characterized by interval along the unit hyperbola.

In terms of speed, rapidity is expressed $\mu = \tanh^{-1}{v/c}$. As $v$ approaches $c$, $\mu$ approaches infinity. So with this natural parameterization, there is no "speed limit".

Update

To get an idea why rapidity is more natural, look at this Wikipedia page and see how simple the expression is for $\gamma$ (simple enough to write it here: $\gamma = \cosh \mu$), and how closely the Lorentz transformation equations resemble the equations for rotation in Euclidean space.

Note also the simplicity of the formula for addition of velocity: $\mu_\mathrm{final} = \mu_1 + \mu_2$

Update #2

This answer gives an excellent real explanation for why rapidity is the natural way to parametrize boosts.

I think the point is, that the Minkowski Metric is actually not a real metric in the strong mathematical sense, because it is not positive definite like the Euclidean scalar product. This introduces restrictions which you don't have with a normal Euclidean Metric.

The Minkowski form is a symmetric bilinear form, but it is not true a scalar product, since it fails to be positive definite, which means the property $\langle v,v\rangle=0\implies v=0$ fails. Calling $Q(x)=\langle x,x\rangle$ the quadratic form associated to the bilinear form, this means $Q$ is not a "norm squared" because it can become zero for nonzero vectors, and indeed for the Minkowski form it can become negative as well. The vectors $v$ with $Q(v)=0$ are called (for some mysterious reason) isotropic vectors, or in case of the Minskowski forn "light vectors", and they form a cone in the vector space. Every linear operator that repects the Minkowski form (Lorentz transformations, which are the orthogonal matrices for this form) must preserve the value of$~Q$, so it can only map isotropic vectors to isotropic vectors. Moreover the isotropic vectors separate the sets of vectors with $Q(v)>0$ and the vectors with $Q(v)<0$, and these sets are also preserved by Lorentz transformations. So the limiting role of light vectors as separating space-like and time-like vectors is a direct consequence of Lorentz transformations respecting the Minkowski form.

For a true scalar product, there is no such thing as nonzero isotropic vectors, so none of the qualitative stuff above applies. The only geometric shapes naturally defined by the metric are spheres. An important difference for the group of orthogonal transformations is that it is "compact" in the case of a true scalar product (meaning that orthogonal transformations cannot "go off to infinity" in any way), which is a consequence of sphere being compact (in contrast to for instance the light cone). This has important mathematical consequences, such as that noe can meaningfully integrate (and therefore average) over the group of orthogonal transformations, which cannot be done forthe group of Lorentz transformations.

The other answers are correct, but I think there is one notion that hasn't yet been explored.

The rotation group is compact. The Lorentz group is noncompact. Take any element $X\in\mathfrak{so}(N)$ from the Lie algebra $\mathfrak{so}(N)$ of the former and $e^{\tau\,X}$ is a periodic function. Not so for the Lorentz group: no finite sequence of finite boosts brings one back to the identity. So we see that the notions of "compact" versus "noncompact" are almost the same question as yours.

So, so far this answer is tautologous - it merely gives us a different way to put your question. But - and here is what I think of as the "answer" part - this compactness versus noncompactness shows up locally as well. A group of operators that conserves the Pythagorean metric needfully has a Lie algebra (a local thing) with the commutation relationships of $\mathfrak{so}(N)$. And the Killing form of these commutation relationships is negative definite: this tells you that the group must be compact. On the other hand, if we do the same thing with a group of operators that conserves the Minkowski pseudo-metric, then the Killing form is indefinite and nondegenerate. This tells us that the Lorentz group cannot be compact, by dint of a theorem that now puts everything into perspective:

Theorem: Let $\mathfrak{G}$ be a Lie group, $\mathfrak{g}$ its Lie algebra and $$\mathscr{K}:\mathfrak{g}\times\mathfrak{g}\to\mathbb{R};\,\mathscr{K}(X,\,Y) = \mathrm{tr}(\mathrm{ad}(X)\,\mathrm{ad}(Y))$$

the Killing form on $\mathfrak{g}$. Let further $\mathfrak{G}$ have either with trivial or discrete center. Then $\mathfrak{G}$ is compact if and only if $\mathscr{K}$ is negative definite.

Proof: see S. Helgason "Differential geometry Lie groups and symmetric spaces" Chap. II, section 6, prop. 6.6.

Since the Minkowski pseudo-metric is nondegenerate, $SO(1,\,3)$ (and, more generally, $SO(p,\,q);\,p,\,q\in\mathbb{N}^+$) has at most a discrete center (actually it has trivial center, but we don't need this). This is because $\mathrm{ad}(Z)=0$ for a central Lie algebra element $Z$ and so $\mathscr{K}(Z,\,\cdot)=0$ and groups with continuous centers must have degenerate Killing forms. Therefore it cannot be compact. Likewise $SO(N)$ has at most discrete center (indeed trivial center) therefore it must be compact.

In Summary: So this is exactly how the nontrivial signature of the Minkowski pseudometric discussed in the other answers weighs on your question: the nontrivial signature makes compactness of the Lorentz group impossible.