ex: ln lim x-> 0+ x^(sin x), y=x^(sin x)

using L'Hopital's Rule, the book says the answer is 1 but i don't know how to get to this answer

Thanks so much!

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- Jul 28th 2009, 06:05 PMriceyL'Hopital's Rule (a different problem)
ex: ln lim x-> 0+ x^(sin x), y=x^(sin x)

using L'Hopital's Rule, the book says the answer is 1 but i don't know how to get to this answer

Thanks so much! - Jul 28th 2009, 06:45 PMANDS!
Looking at the individual functions, we see that both approach 0 as X approaches zero from the left. This would be an indeterminate power, of the form and thus we have to use L'hopital's rule.

To solve this we first take the natural logarithim of our function:

Which then gives us:

This then becomes an indeterminate product of the form

Do you see what to do at this point? - Jul 28th 2009, 06:55 PMricey
it says latex error! syntax error :(

- Jul 28th 2009, 07:02 PMANDS!
Sorry. I suck at this LaTex stuff. Finally figured it out. By accident.

- Jul 28th 2009, 07:16 PMricey
i'm really sorry but how does lny = ln x ^sin x

- Jul 28th 2009, 07:28 PMANDS!
No need to be sorry. Our properties of logarithims says:

is equal to . Because is really ; and hopefully we know a logarithmic property that allows us to manipulate that. - Jul 28th 2009, 07:42 PMricey
hmm okay so first let me ask a question:

does ln lim x sinx = lim ln x sinx ?

and i did see how you got ln y= sin x ln x

from there i can say lim sinx ln x ?

x->0+ - Jul 28th 2009, 08:19 PMVonNemo19
- Jul 28th 2009, 08:37 PMANDS!
Hmmm, I believe he is asking if he can write

is equal to , which is of course no. I'm thinking OP is applying limit rearrangment of the type:

is the same as

The first way of writing it is how we want to set it up.

His second question is can he write it like this:

Which is the proper way to write it, but not using the logic he used. I apologize OP for not writing out the limit notation, as I just assumed you realized where they went.