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x^x as x->0 without L'Hospital
Posted:
Aug 7, 1997 6:20 AM
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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://www-cm.math.uiuc.edu
http://www-cm.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
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