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Math Help - L'Hopital's Rule (a different problem)

  1. #1
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    Question L'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!
    Last edited by ricey; July 28th 2009 at 05:26 PM.
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    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 0^0 and thus we have to use L'hopital's rule.

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

    \ln(y)=\ln(x)^{\sin(x)}

    Which then gives us:

    \ln(y)=\sin(x)\ln(x)

    This then becomes an indeterminate product of the form 0*-\infty

    Do you see what to do at this point?
    Last edited by ANDS!; July 28th 2009 at 06:01 PM.
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  3. #3
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    it says latex error! syntax error
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  4. #4
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    Sorry. I suck at this LaTex stuff. Finally figured it out. By accident.
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  5. #5
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    i'm really sorry but how does lny = ln x ^sin x
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  6. #6
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    No need to be sorry. Our properties of logarithims says:

    \ln(b)^a is equal to a\ln(b). Because \ln(b)^a is really \log_{e}(b)^a; and hopefully we know a logarithmic property that allows us to manipulate that.
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  7. #7
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    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+
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  8. #8
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    Quote Originally Posted by ricey View Post
    hmm okay so first let me ask a question:

    does ln lim x sinx = lim ln x sinx ?
    No. Because this implies that x=\ln{x}. Obviously not true.

    and i did see how you got ln y= sin x ln x
    from there i can say lim sinx ln x ?
    x->0+[/quote]

    I'm not sure that I understand your question...
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  9. #9
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    Hmmm, I believe he is asking if he can write

    \lim_{x \rightarrow 0^{+}} \sin(x) \ln(x) is equal to \sin(x) \lim_{x \rightarrow 0^{+}}\ln(x), which is of course no. I'm thinking OP is applying limit rearrangment of the type:

    \lim_{x \rightarrow a} \sqrt[n]{f(x)} is the same as  \sqrt[n]{\lim_{x \rightarrow a}f(x)}

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

    His second question is can he write it like this:
    \lim_{x \rightarrow 0^{+}} \sin(x) \ln(x)

    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.
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