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Math Help - Limit with Indeterminate Form

  1. #1
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    Limit with Indeterminate Form

    Hi. I have this problem on my homework:

    <br />
\lim_{x \to 1} \frac{1}{ln(x)}-\frac{1}{x-1}<br />

    I know it's the indeterminate form \infty - \infty. I've tried using l'Hopital's Rule, but it doesn't seem to help. (It just becomes \lim_{x \to 1} - \frac{1}{xln(x)^2} - \frac{1}{(x-1)^2} which isn't any better.)
    I've also tried combining the fractions and taking the derivative, which is extraordinarily tedious and also doesn't help. I know by graphing that the limit is supposed to be 1/2, but I can't figure out a way to make that work. Is there something I'm missing? I haven't been this bothered by a math problem in a long time. Any help you can offer would be greatly appreciated. Thanks!
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  2. #2
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    Hello,
    Quote Originally Posted by capandbells View Post
    Hi. I have this problem on my homework:

    <br />
\lim_{x \to 1} \frac{1}{ln(x)}-\frac{1}{x-1}<br />

    I know it's the indeterminate form \infty - \infty. I've tried using l'Hopital's Rule, but it doesn't seem to help. (It just becomes \lim_{x \to 1} - \frac{1}{xln(x)^2} - \frac{1}{(x-1)^2} which isn't any better.)
    I've also tried combining the fractions and taking the derivative, which is extraordinarily tedious and also doesn't help. I know by graphing that the limit is supposed to be 1/2, but I can't figure out a way to make that work. Is there something I'm missing? I haven't been this bothered by a math problem in a long time. Any help you can offer would be greatly appreciated. Thanks!
    You cannot apply l'Hospital's rule in this form


    Let t=x-1

    The limit is now : \lim_{t \to 0} \frac{1}{\ln(1+t)}-\frac 1t

    =\lim_{t \to 0} \frac{t-\ln(1+t)}{t \ln(1+t)}

    now apply the rule twice and you'll get the result (1/2)
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  3. #3
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    Wow, thank you for your quick response! I feel dumb, but I appreciate it.
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