# x^x

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• March 16th 2006, 12:11 PM
Treadstone 71
x^x
How do I show that x^x->1 as x->0?
• March 16th 2006, 12:19 PM
rgep
You could try showing that $x \log x \rightarrow 0$ as $x \rightarrow 0$. If the limit exists it must be zero: consider $x = 2^{-n}$.
• March 16th 2006, 12:49 PM
Treadstone 71
Got it. Used l'Hopitals