Conceptually, the tidal effects of Newtonian gravity differ drastically from that of general relativity. In Newtonian gravity, the effect is due to the change in gravitational force across a certain distance, while in general relativity it corresponds to geodesic deviation and is a result of curvature.
However, from how I see it, they are not very different mathematically.
The Newtonian tidal equation is derived by making a Taylor expansion of the gravitational field.
$$a\approx 2 \Delta r ~G\frac{M}{r^3}$$
The relativistic one uses the geodesic deviation equation and the Riemann curvature tensor for the Schwarzschild metric.
$$a = \Delta r ~\frac{r_s}{r^3}\approx 2 \Delta r ~G\frac{M}{r^3}$$
They both seem to give the same result, even in regions that relativistic descriptions should seem to dominate, like near a black hole. For example, the calculation of the Roche limit of a black hole seems to be identical to that of a more Newtonian object like a star.
Am I missing something here? Am I being too liberal with the "$\approx$" sign? Perhaps there are regimes in which relativistic tidal effects are dramatically different from Newtonian? Or do they differ only conceptually, but in fact produce the same numbers?