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One way this can be explained is from the perspective of numerically approximating an integral. From this perspective, the concordance of "continuous" and "low frequency" has to do with the low frequency regime being where the approximation of the "continuous" integral is valid/best.

If you wanted to numerically compute an approximation to $\int_0^b f(x)dx$ you could use Riemann sums producing $$\int_0^b f(x)dx \approx \Delta x\sum_{n=0}^{\lfloor b/\Delta x\rfloor}f(n\Delta x)$$ The quality of this approximation will depend on the step size, $\Delta x$. The smaller the better.

The integral we care about is improper. Ignoring any analytic subtleties I'll just say that we'll have $$\int_0^\infty f(x) dx \approx \Delta x\sum_{n=0}^\infty f(n\Delta x)$$ assuming $f$ is sufficiently nice and $\Delta x$ sufficiently small.

The specific integrals, referring to section 2.4 of *Elements of Radio Astronomy* which you reference, are the integrals in the average energy, $$\langle E \rangle = \frac{\int_0^\infty EP(E)dE}{\int_0^\infty P(E)dE} \quad \text{where} \quad P(E)\propto e^{-\frac{E}{kT}}$$

Approximating each of the integrals independently produces $$\begin{align}
\int_0^\infty EP(E)dE 
& \approx (\Delta E)^2\sum_{n=0}^\infty n(e^{-\frac{\Delta E}{kT}})^n \\\\
& = (\Delta E)^2 e^{-\frac{\Delta E}{kT}} / (1- e^{-\frac{\Delta E}{kT}})^2
\end{align}$$
and $$\begin{align}
\int_0^\infty P(E)dE 
& \approx \Delta E\sum_{n=0}^\infty (e^{-\frac{\Delta E}{kT}})^n \\\\
& = \Delta E / (1 - e^{-\frac{\Delta E}{kT}})
\end{align}$$
with the resulting ratio being
$$\langle E \rangle \approx \frac{\Delta E }{e^{\frac{\Delta E}{kT}} - 1}$$

We're going to choose $\Delta E = h\nu$ and the quality and thus validity of this approximation will improve as $\nu$ shrinks. Of course, the physics tells us that the approximation goes the other way, i.e. the integral is a continuum approximation of the sum. Regardless of the which expression we consider the approximation, the point is that the expressions are only approximately equivalent for sufficiently small step sizes, i.e. sufficiently small values of $\nu$. Many numerical methods have constraints on how big the step size can be before the method becomes unstable at which point the bounds on errors no longer apply. It is likely the ultraviolet catastrophe can be viewed as this numerical method becoming unstable.