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**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!