Newtonian Limit of Schwarzschild metric The Schwarzschild metric describes the gravity of a spherically symmetric mass $M$ in spherical coordinates:
$$ds^2 =-\left(1-\frac{2GM}{c^2r}\right)c^2 \, dt^2+\left(1-\frac{2GM}{c^2r}\right)^{-1}dr^2+r^2 \,d\Omega^2 \tag{1}$$
Naively, I would expect the classical Newtonian limit to be $\frac{2GM}{c^2r}\ll1$ (Wikipedia seems to agree), which yields
$$ds^2 =-\left(1-\frac{2GM}{c^2r}\right)c^2 \, dt^2+\left(1+\frac{2GM}{c^2r}\right)dr^2+r^2 \,d\Omega^2 \tag{2}$$
However, the correct "Newtonian limit" as can be found for example in Carroll's Lectures, eq.(6.29), is
$$ds^2 =-\left(1-\frac{2GM}{c^2r}\right)c^2 \, dt^2+\left(1+\frac{2GM}{c^2r}\right)\left(dr^2+r^2 \,d\Omega^2\right) \tag{3}$$
Question: Why is the first procedure of obtaining the Newtonian limit from the Schwarzschild solution incorrect?
 A: Consider the change of coordinate
$$r=r'\left(1+{{\cal G}m\over 2r'c^2}\right)^2={r'}^2(1+U)^2$$
where $U={\cal G}m/2r'c^2$. One can check that
$$1-{2{\cal G}m\over rc^2}=1-{2{\cal G}m\over r'c^2(1+U)^2}
={(1-U)^2\over (1+U)^2}$$
Moreover,
$${dr\over dr'}={d\over dr'}\left[r'\left(1+{{\cal G}m\over 2r'c^2}
    \right)^2\right]=(1-U)(1+U)$$                                                    so that
$$dr=(1-U)(1+U)dr'$$
The Schwarzschild metric becomes
$$\eqalign{
    &ds^2=\!c^2\!\left(1-{2{\cal G}m\over rc^2}\right)dt^2
    -\left(1-{2{\cal G}m\over rc^2}\right)^{-1}dr^2
    +r^2d\theta^2+r^2\sin^2\theta d\varphi^2                        \cr
    &=c^2\left(1-{2{\cal G}m\over rc^2}\right)dt^2
    -{(1+U)^2\over (1-U)^2}(1-U)^2(1+U)^2dr'^2
    -{r'}^2(1+U)^4\left[d\theta^2\!+\!\sin^2\theta d\varphi^2\right]\cr
    &=c^2\left(1-{2{\cal G}m\over rc^2}\right)dt^2-(1+U)^4
    \left[d{r'}^2+{r'}^2d\theta^2\!+\!{r'}^2\sin^2\theta
    d\varphi^2\right]                                               \cr
    &=c^2\left(1-{2{\cal G}m\over {r'}c^2}+{\cal O}(U^2)\right)dt^2
    -\big(1+4U+{\cal O}(U^2)\big)\left[d{r'}^2+{r'}^2d\theta^2\!
    +\!{r'}^2\sin^2\theta d\varphi^2\right]                         \cr
    &=c^2\left(1-{2{\cal G}m\over {r'}c^2}\right)dt^2
    -\left(1+{2{\cal G}m\over {r'}c^2}\right)\left[d{r'}^2
    +{r'}^2d\theta^2\!+\!{r'}^2\sin^2\theta d\varphi^2\right]       \cr
    }$$
as expected.
A: Carroll is merely matching the Schwarzschild solution to the linearized weak field solution, treated as a consistent truncated Laurent series in $c^{-1}$, cf. this Phys.SE post. The main point is that the spatial components of the metric are subleading in an $c^{-1}$ expansion and may receive non-trivial contributions in order to maintain EFE.
A: If $\frac{2GM}{c^2R}<<1$ both expressions are valid as approximations.
But the second one presents the expression $dr^2 + r^2 d\Omega^2$ detached. And that is the square of a generic path element in spherical polar coordinates.
Being an elementary spatial path, it can be then replaced by: $dx^2 + dy^2 + dz^2$
