Thread: 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!

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 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!

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Solve Like This :

We Know that :




So

Originally Posted by tonio

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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

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

and is the indeterminate form when ...
After solving this indeterminate you just substitute in the exponential because, as already pointed out, the exp function is a continuous one.

Tonio

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