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Math Help - limit help

  1. #1
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    limit help

    I have the limit as x goes to 1 (1/lnx - x/x-1)
    I am seeing a lot of these limits, with two different fractions. Is a good strategy finding a common denonometer and making it into one fraction? Then use Lhosipatals rule if it applies to find the limit? That is what im trying to do but I am not getting the answer in the book.
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  2. #2
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    \lim_{x\rightarrow{1}}[\frac{1}{ln(x)}-\frac{x}{x-1}]

    Rewrite as:

    \lim_{x\rightarrow{1}}\frac{-xln(x)+x-1}{xln(x)-ln(x)}

    L'Hopital gives:

    \lim_{x\rightarrow{1}}\frac{-xln(x)}{x-1+xln(x)}

    L'Hopital again:

    \lim_{x\rightarrow{1}}\frac{-ln(x)-1}{ln(x)+2}=\frac{-1}{2}
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  3. #3
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    Quote Originally Posted by galactus View Post
    \lim_{x\rightarrow{1}}[\frac{1}{ln(x)}-\frac{x}{x-1}]

    Rewrite as:

    \lim_{x\rightarrow{1}}\frac{-xln(x)+x-1}{xln(x)-ln(x)}

    L'Hopital gives:

    \lim_{x\rightarrow{1}}\frac{-xln(x)}{x-1+xln(x)}

    L'Hopital again:

    \lim_{x\rightarrow{1}}\frac{-ln(x)-1}{ln(x)+2}=\frac{-1}{2}
    How did you rewrite that fraction? I dont see that.
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  4. #4
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    That's just what you would do if you were subtracting. Basic algebra.

    \frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}
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  5. #5
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    Quote Originally Posted by galactus View Post
    \lim_{x\rightarrow{1}}[\frac{1}{ln(x)}-\frac{x}{x-1}]

    Rewrite as:

    \lim_{x\rightarrow{1}}\frac{-xln(x)+x-1}{xln(x)-ln(x)}

    L'Hopital gives:

    \lim_{x\rightarrow{1}}\frac{-xln(x)}{x-1+xln(x)}

    L'Hopital again:

    \lim_{x\rightarrow{1}}\frac{-ln(x)-1}{ln(x)+2}=\frac{-1}{2}
    When you take the derivative of (-xln(x) + x - 1)/xln(x) - ln(x) I am getting
    (-x - ln(x))/x+ln(x)-(1/x)
    What am i doing wrong? EDIT NVM i forgot a 1
    lol.
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  6. #6
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    Yeah, the biggest trouble with calc is the algebra. It can be tricky and easy to 'flub up'.
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