1. Indeterminate powers

Find the limit:

lim x -> 0+ x^x

write as an exponential:

x^x = e^(xlnx)

I know: lim x -> 0+ xlnx = 0 (we did that earlier with the l'Hospital's Rule)

so,
lim x -> 0+ x^x = lim x -> 0+ e^(xlnx)
= e^0
= 1

2 things I do not understand.
First, how do I know that this is indeterminate?

Second,
lim x -> 0+ x^x = lim x -> 0+ e^(xlnx)
= e^0
Why do I take the limit of the exponent of e if the limit of the entire term is needed?
Thus, is lim x -> 0+ e^(xlnx) the same as e^lim x -> 0+ (xlnx) ??

Thanks!

2. Originally Posted by DBA
Find the limit:

lim x -> 0+ x^x

write as an exponential:

x^x = e^(xlnx)

I know: lim x -> 0+ xlnx = 0 (we did that earlier with the l'Hospital's Rule)

so,
lim x -> 0+ x^x = lim x -> 0+ e^(xlnx)
= e^0
= 1

2 things I do not understand.
First, how do I know that this is indeterminate?

What or who is indeterminate? You said you already evaluated the limit $\displaystyle \lim_{x\to\infty}x\ln x$ by L'Hospital, and you did so because this is, of course, and indeterminate...

Second,
lim x -> 0+ x^x = lim x -> 0+ e^(xlnx)
= e^0
Why do I take the limit of the exponent of e if the limit of the entire term is needed?
Thus, is lim x -> 0+ e^(xlnx) the same as e^lim x -> 0+ (xlnx) ??

Indeed, and you can do this because the exponential function is continuous EVERYWHERE and thus evaluating a limit is, simply, substituing...

Tonio

Thanks!
.

3. Solve Like This :

We Know that :

$\displaystyle x =e ^ {\ln x}$

So

$\displaystyle \lim_{x\rightarrow 0}e^{{\ln x}^x}= \lim_{x\rightarrow 0}e^{x\ln x}=e ^{\lim_{x\rightarrow 0}x{\ln x}}=e ^{\lim_{x\rightarrow 0}\frac {\ln x}{\frac {1}{x}}}\rightarrow {Lopital} = e^0 = 1$

4. Originally Posted by tonio
.
What I did not understand is how do I know that

lim x -> 0+ x^x is indeterminate.

Normally I would check the limit of the term and the exponent...

lim x -> 0+ x = ? and lim x -> 0+ x = ? Then I would get a type like 0/0 or infinite/infinite. If I have a type like that than I am allowed to use the l'Hospital's rule.

In this case I wasn't sure if I can use this check too? So, my question is: What indeterminate type do I have for lim x -> 0+ x^x

5. Originally Posted by DBA
What I did not understand is how do I know that

lim x -> 0+ x^x is indeterminate.

Normally I would check the limit of the term and the exponent...

lim x -> 0+ x = ? and lim x -> 0+ x = ? Then I would get a type like 0/0 or infinite/infinite. If I have a type like that than I am allowed to use the l'Hospital's rule.

In this case I wasn't sure if I can use this check too? So, my question is: What indeterminate type do I have for lim x -> 0+ x^x

$\displaystyle x^x=e^{x\ln x}$ and $\displaystyle x\ln x = \frac{\ln x}{1\slash x}$ is the indeterminate form when $\displaystyle x\rightarrow 0$ ...
After solving this indeterminate you just substitute in the exponential because, as already pointed out, the exp function is a continuous one.

Tonio

6. Originally Posted by DBA
Find the limit:

lim x -> 0+ x^x

write as an exponential:

x^x = e^(xlnx)

I know: lim x -> 0+ xlnx = 0 (we did that earlier with the l'Hospital's Rule)

so,
lim x -> 0+ x^x = lim x -> 0+ e^(xlnx)
= e^0
= 1

2 things I do not understand.
First, how do I know that this is indeterminate?
http://www.math.fsu.edu/~bellenot/class/f99/cal1/indeterminate.pdf