= Solution
Use <Minkowski spacetime> coordinates $x^\mu=(ct,x,y,z)$ and the <Minkowski metric> of signature $(+,-,-,-)$. For a massive particle with three-velocity $\mathbf v=d\mathbf x/dt$ and $|\mathbf v|<c$, <proper time> satisfies
$$
c^2\,d\tau^2=c^2\,dt^2-|d\mathbf x|^2,\qquad
\frac{dt}{d\tau}=\gamma=\left(1-\frac{|\mathbf v|^2}{c^2}\right)^{-1/2}.
$$
Thus the <four-velocity> components are
$$
\boxed{U^\mu=(\gamma c,\gamma v_x,\gamma v_y,\gamma v_z)}.
$$
Here upper indices indicate components in the chosen coordinates; lowering with the <Minkowski metric> changes the signs of the spatial components. Its invariant product defined by the <Minkowski metric> is
$$
U\cdot U=(U^0)^2-\sum_{j=1}^3(U^j)^2
=\gamma^2(c^2-|\mathbf v|^2)=c^2.
$$
The <four-momentum> is $p^\mu=(E/c,\mathbf p)=mU^\mu$, so $E=\gamma mc^2$ and $\mathbf p=\gamma m\mathbf v$. The <Taylor expansion> of the <Lorentz factor> at small $|\mathbf v|/c$ is
$$
\gamma=1+\frac{|\mathbf v|^2}{2c^2}+O(|\mathbf v|^4/c^4).
$$
Consequently $\mathbf p=m\mathbf v+O(m|\mathbf v|^3/c^2)$, recovering ordinary <momentum>. At rest the <four-momentum> has energy $E_0=mc^2$, so subtracting this <rest energy> gives the <kinetic energy>
$$
\boxed{K=E-E_0=(\gamma-1)mc^2
=\frac12m|\mathbf v|^2+O(m|\mathbf v|^4/c^2)}.
$$
Thus the nonrelativistic limit preserves the rest-energy offset as well as the usual <kinetic energy>.
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