$1/f$ noise: is it limited? Basically my question is:

I have this doubt because in contrast to the limit $f \to \infty$ (which is an idealization because all system behaves as a low pass for enough $f$), we can perfectly build systems that not only approach $f \to 0$ but implement $f=0$, for example any DC circuit. And if $1/f$ wouldn't be limited then the energy would be infinite...
If it is limited, at which $f$ and how it decays?
 A: As a matter of principle, the $1/f$ climb needs to be limited at both ends. To quote Wikipedia,

In pink noise, each octave (halving/doubling in frequency) carries an equal amount of noise energy

which means that each frequency interval $[2^{n}f,2^{n+1}f]$ (for $n$ both positive and negative) carries a constant amount of energy. Since there are an infinite number of such intervals between the 1 Hz edge of your diagram and the $\frac{1}{\infty}\:$Hz edge of possible frequencies, a $1/f$ climb that continued indefinitely to the left would carry infinite energy.
In practice, an unlimited climb towards the left would require an infinite measurement time, since it involves arbitrarily small frequencies, and you don't have that much time for your experiment. Thus, in practice, the low-frequency climb of the noise curve will be limited (at most) by the total time taken by your experiment.
A: It's limited to zero, at zero.  First imagine that you have a perfect filter for frequencies from zero to 1 Hz.  You can use your scope to approximate the voltage in that bandwidth.  Now square the voltage and you can specify the noise in volts squared per hertz, as noise is usually specified.  Imagine you retune your filter from zero to 1/2 Hertz.  Half the noise.  Now zero to 1/4 Hertz.  One-quarter the noise.  And so on down until there is no bandwidth left, and no noise.
