Expert: Paul Klarreich - 3/5/2010

Question
Limit x->0 x^x
well by plugging in it's 0^0 undetermined using l'hopital  
d/dx limit x->0 x^x =limit x->0 x^x(ln(x)+1)
doing it again d/dx limit x->0 x^x(ln(x)+1) =  x^x(ln(x)+1)^2 + x^(x-1)
doing it again... d/dx x^x(ln(x)+1)^2 + x^(x-1) will still have x^x and will pretty sure be much more complicated so how can I do an undetermined if taking the derivative makes it harder

Answer
Questioner: hamad
Country: Kuwait
Category: Advanced Math
Private: No
Subject: undetermined form
Question: Limit x->0 x^x
well by plugging in it's 0^0 undetermined using l'hopital  
d/dx limit x->0 x^x =limit x->0 x^x(ln(x)+1)
doing it again d/dx limit x->0 x^x(ln(x)+1) =  x^x(ln(x)+1)^2 + x^(x-1)
doing it again... d/dx x^x(ln(x)+1)^2 + x^(x-1) will still have x^x and will pretty sure be much more complicated so how can I do an undetermined if taking the derivative makes it harder

........................
Yes, 0^0 sure is INdeterminate. (Not UN...)

Do this, any time you have an exponential function:

y = x^x, to find lim y.

ln y = x ln x, and find  lim  ln y.

x ln x HAS to be made a fraction, to use l'Hospital's rule.  So you just make it a fraction, whether you like it or not.

           ln x
x ln x =  --------
           1/x

Now diff:
   1/x
---------- =
  -1/x^2
 x
----- ---> 0.
-1

So ln y --> 0, then y --> e^0 = 1.

That is it.

Not as cute as the other one, but we'll take it.