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 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!