/sci/ - Science & Math

Back at my home computer, and with Mathematica at my side, let me try to sketch this out.

First:
[eqn] x^{\sin x^x} = \exp\left(\sin\left(e^{x \ln x}\right) \ln x \right)[/eqn]
Then we go:
[eqn] \lim_{x\to-\infty} \left( \exp\left(\sin\left(e^{x \ln x}\right) \ln x \right) \right) = \\
\exp\left( \lim_{x\to-\infty} \left( \sin\left(e^{x \ln x}\right) \ln x \right) \right) =\\
\exp\left( \lim_{x\to-\infty} \left( \sin\left(e^{x \ln x}\right) \right)\cdot \lim_{x\to-\infty} \left( \ln x \right)\right) = \\
\exp\left( \sin\left(\lim_{x\to-\infty} \left(e^{x \ln x}\right) \right)\cdot \lim_{x\to-\infty} \left( \ln x \right)\right) = \\
\exp\left( \sin\left( \exp\left( \lim_{x\to-\infty} x \ln x\right) \right) \cdot \lim_{x\to-\infty} \left( \ln x \right)\right) = \\
\exp\left( \sin\left( \exp\left( \lim_{x\to-\infty} \left( x \right) \lim_{x\to-\infty} \left( \ln x \right) \right) \right) \cdot \lim_{x\to-\infty} \left( \ln x \right)\right) = \\
\exp\left( \sin\left( \exp\left( \lim_{x\to-\infty} \left( x \right)\right)^{\lim_{x\to-\infty} \left( \ln x \right)} \right) \right)^{\lim_{x\to-\infty} \left( \ln x \right)}
[/eqn]
Now we only have to calculate two actual limits:
[eqn]
\lim_{x\to-\infty} x = -\infty \\
\lim_{x\to-\infty} \ln x = \lim_{x\to-\infty} i\pi + \ln (-x) = \infty
[/eqn]
Plugging those back in:
[eqn]
\exp\left( \sin\left( \exp\left( -\infty \right)^{\infty} \right) \right)^{\infty}
[/eqn]
And now collapse it inward:
[eqn]
e^{-\infty} = 0 \\
0^\infty = 0 \\
\sin 0 = 0 \\
e^0 = 1
[/eqn]
The problem is that the last step, [math] 1^\infty [/math] is an indeterminate expression. Mathematica will evaluate [math] \lim_{x\to\infty} 1^x [/math] as 1, but that's actually subtly wrong. See http://math.stackexchange.com/questions/10490/why-is-1-infty-considered-to-be-an-indeterminate-form for an explanation.