= Solution
Hold $p=m\gamma$ fixed. At a point with $x\ne t$,
$$
A=\frac{p}{2\gamma\sqrt{\gamma^2(x-vt)^2+y^2+z^2}}
\longrightarrow0,
$$
but
$$
\boxed{\gamma^2A\longrightarrow\frac{p}{2|x-t|}.}
$$
The small-$A$ expansions are
$$
f(A)=1+4A+O(A^2),
\qquad
g(A)=8A+O(A^2).
$$
Consequently, with $H=4p/|x-t|$, the <Ultrarelativistic limit of the boosted Schwarzschild metric away from the null plane> is
$$
\boxed{ds^2=-dt^2+dx^2+dy^2+dz^2+H(dt-dx)^2,\qquad x\ne t.}
$$
Equivalently,
$$
\boxed{g_{tt}=-1+H,\quad g_{tx}=g_{xt}=-H,\quad g_{xx}=1+H,\quad g_{yy}=g_{zz}=1,}
$$
with all other components zero.
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