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Math Help - Limit x-> infinity

  1. #1
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    Limit x-> infinity

    Hi everyone. Got stuck again. Solved it for when x-> infinity but cant seem to get it for when x->0. Appreciate any help at all!

    lim <br />
    \frac{3^{x}+ln{(absx)}}{x^5+x^4}<br />
    x->0

    Thanks!
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  2. #2
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    thanks Dan! But is there a way without using that rule?

    Johan
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  3. #3
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    I thought for L'Hopital's you need both the top and bottom to tend to 0 as x tends to 0. logx tends to -infinity doesn't it?
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  4. #4
    Eater of Worlds
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    You could try expanding and see what you come up with. Sometimes it makes it easier to see. Since x-->0, you shouldn't have to worry about an absolute value.

    \lim_{x\rightarrow{0}}\frac{3^{x}+ln(x)}{x^{5}+x^{  4}}

    =\lim_{x\rightarrow{0}}\left[\frac{ln(x)}{x+1}-\frac{ln(x)}{x}+\frac{ln(x)}{x^{2}}-\frac{ln(x)}{x^{3}}+\frac{ln(x)}{x^{4}}+\frac{3^{x  }}{x+1}-\frac{3^{x}}{x}+\frac{3^{x}}{x^{2}}-\frac{3^{x}}{x^{3}}+\frac{3^{x}}{x^{4}}\right]
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Thomas154321 View Post
    I thought for L'Hopital's you need both the top and bottom to tend to 0 as x tends to 0. logx tends to -infinity doesn't it?
    It can also be infinity over infinity. However that isn't the case here either. Thanks for the catch.

    -Dan
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  6. #6
    GAMMA Mathematics
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    What is the range of ln(x) ? This is a question I never raised nor answered.
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  7. #7
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    The range of lnx is the domain of e^x, ie all real values.
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  8. #8
    GAMMA Mathematics
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    Quote Originally Posted by Thomas154321 View Post
    The range of lnx is the domain of e^x, ie all real values.
    I suppose so; it just grows incredibly slow!
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