Heat resistance of sphere I have been taught that the heat resistance of a hollow cylinder is
R(hollow cylinder) is proportional to ln(outer radius/inner radius)
It was assumed in the derivation that the inner surface is at some temperature t and has some radius r. Steady State and no heat generation also follows. The outer surface is at some temperature T and radius R.
All I am trying to do here is limiting r to zero.
Why does a complete cylinder have infinite heat resistance?
 A: 
I have been taught that the heat resistance of a hollow sphere is: R(hollow sphere) is proportional to ln(outer radius/inner radius)

Well, this isn't correct; as derived in many textbooks (e.g., Incropera & DeWitt) and noted in many handbooks, the thermal resistance for steady-state conduction through a hollow sphere with uniform temperatures on the inside and outside surfaces (of radii $r_1$ and $r_2$, respectively) is
$$R_\text{sphere}=\frac{1}{4\pi k}\left(\frac{1}{r_1}-\frac{1}{r_2}\right),$$
where $k$ is the thermal conductivity.
(The thermal resistance of a cylinder includes a $\ln(r_2/r_1)$ term; perhaps you were thinking of that?)
In any case, we find the inner radius $r_1$ appearing in the denominator for both geometries. How is it that the thermal resistance of these objects (with radial symmetry) can increase to infinity as the internal cavity shrinks?
Because the area $A$ over which the temperature is applied drops to zero. Therefore, as mediated by Fourier's law of conduction $q=-kA\frac{dT}{dx}$, the temperature difference $dT$ required to achieve a given heat flux $q$ tends to infinity; this is equivalent to an infinite thermal resistance.
The situation is broadly similar to trying to heat up an object by heat conduction through a safety pin, say, touching the object's surface. The kinetics of the process are limited by the very small cross section of the pin, and you'd need an enormous temperature at the other end to push through appreciable heat.
