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x^x as x>0 without L'Hospital
Posted:
Aug 7, 1997 6:20 AM


In Calculus&Mathematica, we establish (in the very first lesson before the derivative is introduced) that exponential growth dominates power growth as the variable gets big.
Once you have this, then you can see what x^x does as x>0
Look at Log[x^x] = x Log[x] as x>0 Put x = E^(t) and note x>0 means t>infinity.
So now you have E^(t) Log[E^(t)] = t/E^(t) >0 as t>infinity. (because exponential growth dominates power growth)
Consequently Log[x^x] > 0 as x > 0 Consequently x^x  > 1 as x >0.
Jerry Uhl
 Jerry Uhl juhl@ncsa.uiuc.edu Professor of Mathematics 1409 West Green Street University of Illinois Urbana,Illinois 61801 Calculus&Mathematica Development Team http://wwwcm.math.uiuc.edu http://wwwcm.math.uiuc.edu/dep
[If] logic is the hygiene of the mathematician, it is not his source of food. Andre Weil
Only professional mathematicians learn anything from proofs. Other people learn from explanations. R. P. Boas



