I believe that the Roche limit would be roughly the same as it is on earth. However, the gravity constant would be the same. So for something as massive as a person, you should not experience any destructive forces until you reached extremely close to the object. Probably measured in a handful of miles or less.

As usual the answer is - it depends, but we could estimate forces in question. Assuming my math is correct, at 67 kilometers getting 1 meter closer will results in gravitational pull difference of 137kg. At 40 kilometers, this number increases to ~623kg. I am assuming bad things will happen somewhere between those two ranges.

F = G * (m1 * m2) / r^2 . G = 6.674 / 10^11 (m^3 / (kg * s^2)) | m1 = 5 * 5.972 * 10^24 kg | m2 = 100 kg

Only r in the equation above is not a constant, so we can compute F(r) - F(r+1) to get a difference in force that being 1 meter closer to the black hole would make at a distance r. Basically, this is the difference between how hard your toes would be pulled down vs your heard if you curl into a ball.

  Distance Force          Force + 1      Diff in kg
  1        1.99286E+17    4.98214E+16    1.49464E+16
  1000     1.99286E+11    1.98888E+11    39797421.92
  10000    1992856400     1992457888     39851.15023
  20000    498214100      498164282.3    4981.767364
  30000    221428488.9    221413727.7    1476.11612
  40000    124553525      124547297.6    622.744272
  50000    79714256       79711067.53    318.8474585
  60000    55357122.22    55355277.03    184.5191277
  70000    40670538.78    40669376.79    116.1990494

This is the premise of a book by Thomas Wren:

https://www.goodreads.com/book/show/2335064.The_Doomsday_Eff...

From (distant) memory it was a fun read with some science but not overly realistic wrt to how humanity would react.

My understanding is that a black hole large enough to be colder than the CMB will grow (gain more mass energy by absorbing CMB photons, even in "empty" space, than it loses by Hawking radiation), continuing to cool.

I assume but am not sure that the cooling of an already-sufficiently-cold black hole in this manner is slower than the cooling of the CMB itself due to expansion, so that at some point as the universe winds down it will begin to shrink again and eventually evaporate, but definitely over the lifetime of any star we see a black hole this side won't radiate significantly, or evaporate.

Not quite. Point singularities can't spin, and since a black hole with no spin as about as possible as real mass with a temperature of absolute zero, every black hole we're ever likely to observe is actually a ringularity or other two-dimensional configuration.

As I understand it, the fact that black holes are rotating precludes them from actually being a point singularity -- "rotation" having no meaning for a non-dimensional point. So they need to have some kind of disk or ring structure (still infinitely thin, however). (Note: I'm not a physicist, could be completely wrong).

"when you stare into the abyss..."

This is absolutely something you can do with ASCII dot-matrix printers, and it's not even malformed! It was common to end a line with a CR rather than a CRLF in order to overstrike some characters three times for bold, or overstrike them with _ for underlining; somewhat less common was to overstrike several different characters to get a solid block (such as 0, #, and *). (Accented characters were likewise achieved by overstriking `, ', and ~, but usually using backspace rather than CR.) With a normal amount of ink on the ribbon, you would punch through the paper after less than ten overstrikes.

I understand that line printers (those that printed a line at a time rather than a character at a time) also supported overstrike sequences, but I never punched through the paper on one. Likewise, presumably the same approach would work with a daisy-wheel or typeball printer, but I can't testify.

And absorbed by black hole would pretty memorable at my memorial. Its win-win.