# Indefinite forms and l'hopital

1. Feb 14, 2012

### Square1

1. The problem statement, all variables and given/known data
*indeterminate* oops

the limit of x^x as x goes to zero from the right

2. Relevant equations

Going to be using L'hopital, and related algebraic manipulations to convert to indefinite form 0/0, infinity/infinity

3. The attempt at a solution

My understanding is that this limit produces initially by plugging in 0 the indefinite exponential form 0^0, and so I have the choice to either:

a) take natural log, bring down the x...
b) write in exponential form of 'e'

I always use the log method bc I dont really get truly whats going on in method b. When I use 'a', I end up with the form -x, which gives me a limit of 0. When 'b' is used, the textbook says the limit is 1, because 'e'^xlnx = 'e'^0 = 1. This is in Stewart's single var. calc text as an exponential indefinite form example.

Thanks.

Last edited: Feb 14, 2012
2. Feb 14, 2012

### HallsofIvy

Staff Emeritus
(a) and (b) are essentially the same. Any number, u, can be written as $e^{ln(u)}$ so that $x^x= e^{ln(x^x)}= e^{xln(x)}$. Since $e^x$ is continuous, $lim e^{xln(x)}= e^{\lim (x ln(x)}$.

3. Feb 14, 2012

### Square1

Ok but the book gets an answer of one, and I keep getting zero!

4. Feb 14, 2012

### Staff: Mentor

You might be getting the limit of the ln of your expression. If this limit is 0, then the limit of the expression itself (rather than its ln) will be 1.

Note that under certain conditions, the operations "limit of ..." and "ln of ..." can be switched.

5. Feb 14, 2012

### Square1

Ok I got it! I think it is what mark was talking about.

I evaluated the limit of xlnx = 0, and forgot that by manipulating theoriginal expression to be in terms of ln, that I have lny on the other side. So I forgot to do the last step which is solve for y in lny = 0 , which equals 1.