Indefinite forms and l'hopital

[Square1]

Homework Statement

*indeterminate* oops

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

Homework Equations

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

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 don't really get truly what's 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.

 

[HallsofIvy]

(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, $lim e^{xln(x)}= e^{\lim (x ln(x)}$.

 

[Square1]

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

 

[Mark44]

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

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.

 

[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.