0^0 = 1?

[Thomas Okken]

Well, yes, y^x is a bivariate function, so you can come at (0, 0) from more than just two directions. If we parameterize the direction as an angle φ from the positive x axis, we get

y^x = (r sin(φ))^(r cos(φ))

and the value we're looking for is the limit of that as r goes to zero.
For all φ ∈ (0, π), so positive y, this limit is 1; for all φ ∈ (π, 2π), so negative y, this limit is also 1, albeit with a side trip through the complex plane.

This leaves only the cases φ = 0 and φ = π, i.e. y = 0 with x > 0 and x < 0, respectively. Lim (x->0) 0^x is undefined if you come at it from below, because of division by 0. From above it's tricky, because while 0^x = 0 for all x > 0, suggesting simply continuing this line yielding 0^0 = 0, its behavior at x = 0 is rather special:

0^x = e^(x * log(0)) = e^(x * -Inf)

because x * -Inf = -Inf for all x > 0, and undefined when x = 0. The usual tricks for resolving limits by dropping higher-order terms or cancelling factors don't work here because that infinity really is a hard infinity, not a factor that goes to inifinity as we approach the limit.

So, lim (r->0) (r sin(φ))^(r cos(φ)) = 1 for φ ∈ (0, π) ∪ (π, 2π) and undefined for φ = 0 and φ = π.

Make of that what you will. :-)