= Solution
At $(x,0,0)$ the field is perpendicular to the loop plane. Its Biot-Savart numerator is
$$
\left[d\boldsymbol\ell'\times(x-x',-y',0)\right]_z
=-x\,dy'+x'\,dy'-y'\,dx'.
$$
The leading denominator variation must be kept:
$$
\frac1{[(x-x')^2+y'^2]^{3/2}}
=\frac1{|x|^3}\left(1+\frac{3x'}x+\cdots\right).
$$
The integral of $-x\,dy'$ is zero, but its product with $3x'/x$ is not. Keeping all terms of the leading nonzero order gives
$$
B_z\simeq\frac{\mu_0I}{4\pi|x|^3}
\oint[-2x'\,dy'-y'\,dx'].
$$
For the counterclockwise rectangle, $\oint x'\,dy'=\Delta x\Delta y$ and $\oint y'\,dx'=-\Delta x\Delta y$. Hence
$$
\boxed{\mathbf B(x,0,0)\simeq-\frac{\mu_0\mathbf m}{4\pi|x|^3}.}
$$
This transverse field is opposite to the <magnetic dipole moment>. Replacing every denominator by a constant before examining the term proportional to $x$ would give the wrong answer.
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