Me and my friends enjoy setting each other puzzles, and one that I have come up with recently is the following:
$$\lim_{x\to \infty} \sqrt[x]{-1}$$
My idea is that by thinking of rooting something as rotation in the complex plane (where you are halving the angle in polar coordinates with each root), you will find that the answer converges to 1.
Or more formally:
$$ -1 = e^{i\pi} $$ $$\sqrt[x]{-1} = \sqrt[x]{e^{i\pi}} $$ $$ = (e^{i\pi})^{1/x}$$ then as x tends to infinity, the answer tends to 1. Pretty neat, right?
What I'm wondering though is whether there is a second answer - as in the complex plane there is normally a second root opposite to the first. Obviously this opposite route would tend towards -1 just as the above sum tends to 1.
So is the answer 1, -1, both, or undefined?
I would go for undefined (or the unit circle), since $-1=e^{i(\pi + 2\pi k)}$, $k\in {\Bbb Z}$ so that $(-1)^{1/x}= e^{i(\pi + 2\pi k)/x}$ which for suitable choice of $k=k(x)$ may accumulate on any point on the unit circle when letting $x\rightarrow \infty$.