how fast can you sweep sinewave
To ask "how fast can you sweep a sinewave?" is to ask "How fast can you hit $\frac{F_s}{2}$?".

Because, sweeping a sinewave is equivalent to modulating it. 

A sine wave looks like:
$$x(n) = \sin(n 2 \pi f / F_s)$$

Let's focus on that instantaneous phase. I have put $n$ first, on purpose, to show that phase, in discrete time, is a line that departs from $0$ at a slope of $\frac{2 \pi f}{F_s}$. Even more descriptively, $2 \pi \frac{f}{F_s}$.

The sharpest slope for a given $F_s$ that can exist is at $\frac{2 \pi \frac{F_s}{2}}{F_s} = \frac{2 \pi F_s}{2 F_s} = \pi$.

Anything even slightly higher than that rate will start "catching" the sinusoid at its "past", or in other words, create aliasing.

So, how fast can you sweep a sinusoid?

Consider a cartesian plane. It's $x$ axis corresponds to time and it's $y$ axis corresponds to rate of phase change. We mark a red line on the $y$ axis at $\pi$. That's our limit. We can't cross that. The time axis changes in terms of $T_s = \frac{1}{F_s}$. So, theoretically, we could go $0 \ldots \frac{F_s}{2}$ in just one step, in just one $T_s$.

Is that going to look like a sweep? 

It is going to look like a sweep, as much as a sinusoid looks like a sinusoid when it is sampled at some $\frac{F_s}{2}$. That's two samples per period and it looks exactly like a triangle and nothing like a sinusoid, but when you are sampling at $F_s$, that's the best sinusoid you can "see" at $\frac{F_s}{2}$.

This cartesian plane representation though, helps convey another point too. That is, how many frequencies did the sweep end up going through anyway? In the extreme example given above, instantaneous phase was so rapidly modified that effectively just two frequencies showed up, 0 and $\frac{F_s}{2}$. Sometimes, when you are trying to obtain the frequency response of a system, you might "miss" a part of it because your sweep is so fast that it "jumps" over the mode you should have excited.

Therefore, a more meaningful way to determine your sweep rate is to accept an average power per Hz (or frequency band) that you want your chirp to deliver and then find the rate of phase change that "allows enough samples" to deliver that power. 

Hope this helps.