How far into the future can we go by traveling close to a black hole? If we sent someone on a path that passed as near as possible to a black hole without getting pulled in, how far into the future would they go due to time dilation?  Let's assume the black hole is 5 solar masses (I am assuming the mass will affect the calculation).  I would like to know from two considerations:


An observer that can survive anything (e.g. ideal case)
The forces that a human can tolerate (e.g. a human would not be able to get as close to a black hole as would be ideal)
The way that you have specified the question, the answer is as far as you like. You simply put your spaceship into any orbit around the black hole and wait.

A more sensible question is what is the largest time dilation factor that can be accomplished - i.e. that maximises your travel time into the future for a given amount of proper time experienced on the spaceship. 

This in turn is governed by how close to the black hole you can come and still tolerate the tidal forces. If you don't put a limit on this (your first case), then the answer is again infinite; you can hover as close to the event horizon as you like, using an enormous amount of rocket fuel, and the time dilation (see below) can be arbitrarily large.

Your second case is more realistic. Roughly we can say that the tidal acceleration across a body of length $l$ is given by $GMl/4r^3$, where $M$ is the black hole mass and $r$ is the distance from the black hole. If we make this acceleration equal to say $1 g$, and your body length $l \sim 1$m, then for a $5M_{\odot}$ black hole $r \simeq 2500$ km (well outside the Schwarzschild radius of 15 km).

If you could "hover" at this radius, then the time dilation factor would be
$$\frac{\tau}{\tau_0} = \left( 1 - \frac{2GM}{rc^2}\right)^{1/2},$$
where $\tau$ is the time interval on a clock on the spaceship and $\tau_0$ is the time interval well away from the black hole.

For $M=5M_{\odot}$ and $r = 2500$ km, this factor is 0.9970.

If the spaceship is in a circular orbit, the factor is $(1 - 3GM/rc^2)^{1/2} = 0.9955$.

These factors are perhaps not as big as you might have imagined!