How small is "vanishingly small" when working with the Schwarzschild metric? In Exploring Black Holes, when dealing with the metric for polar coordinates for flat spacetime, he says "This derivation is valid only when dΘ is small - vanishingly small in the calculus sense - so that the differential segment of arc rdΘ is indistinguishable from a straight line." (page 3-2)
So just how small is vanishingly small?  Lets say we are working at r = 1,000 metres from the event horizon.  Can we reasonable work with a line 1 degree wide? or .1 degree? or .01 degree?
 A: Your question pertains to the 2D Euclidean metric in polar coordinates.  Given that
$$x = r \cos \phi \qquad y = r \sin\phi$$
It's easy to show that the distance between any two points $(r_1,\phi_1)$ and $(r_2,\phi_2)$ is
$$\Delta s^2 = \Delta x^2 + \Delta y^2 =\big(r_2\cos(\phi_2) - r_1\cos(\phi_1)\big)^2 + \big(r_2\sin(\phi_2) - r_1\sin(\phi_1)\big)^2$$
$$= r_1^2 + r_2^2 - 2 r_1 r_2\cos\big(\phi_2 - \phi_1\big)$$
If we define $\Delta r \equiv r_2-r_1$ and $\Delta \phi\equiv\phi_2-\phi_1$ and drop the subscript on $r_1$, then
$$\Delta s^2 = \Delta r^2 + 2r^2\big(1-\cos(\Delta \phi)\big) + 2 r\Delta r \big(1-\cos(\Delta \phi)\big)$$
This expression is exact for any value of $\Delta r$ and $\Delta \phi$.  However, expanding about $\Delta \phi,\Delta r= 0$, this becomes
$$\Delta s^2 = \Delta r^2 + r^2 \Delta \phi^2 + \left[  r\Delta r  (\Delta \phi)^2 -r^2 \frac{(\Delta \phi)^4}{12} +\ldots\right]$$
The terms outside of the brackets are quadratic in $\Delta r$ and $\Delta \phi$, whereas the terms inside the brackets are cubic and quartic, respectively. If you want to know how good a particular approximation is, then you can evaluate the terms in the bracket and compare them to the terms outside.
Note that we can also rewrite the above expression as
$$\Delta s^2 = \Delta r^2 + r^2 \Delta \phi^2 \left[ 1 + \frac{\Delta r}{r} + \frac{(\Delta \phi)^2}{12} + \ldots \right]$$

The larger point is that if we let $\Delta r,\Delta \phi$ be infinitesimal quantities $\mathrm dr,\mathrm d\phi$, then lowest order terms the quadratic ones and we find that
$$\mathrm ds^2 = \mathrm dr^2 + r^2 \mathrm d\phi^2$$
This can be used to compute the distances along any curve $C :[0,1]\ni \lambda \mapsto \big(r(\lambda),\phi(\lambda)\big)$ by integration:
$$\Delta s = \int_{0}^{1} \sqrt{r'(\lambda)^2 + r(\lambda)^2\phi'(\lambda)^2 } \mathrm d\lambda$$
