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Math Help - L'Hopital's Rule?

  1. #1
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    L'Hopital's Rule?

    Can I get an answer using L'Hopital's rule here?

    Lim x --->2 \frac{x^2-4}{ln(x^2-3)}

    I know that they'll \frac{0}{0}, but I just wasn't sure. Can I get some help to work through this one?
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  2. #2
    Super Member wingless's Avatar
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    \lim_{x\to 2 } \frac{x^2-4}{\ln(x^2-3)}

    \frac{2^2-4}{\ln(2^2-3)}

    \frac{4-4}{\ln(1)}

    \frac{0}{0}

    Yes, you can use L'hopital here.
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  3. #3
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    L'hŰpital

    You will have no problem using l'hŰpital there, the only problem with l'hŰpital could be that you get nowhere by using derivatives, but this will not be the case
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  4. #4
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    \frac{2x}{\frac{1}{x^2-3}}

    plugging in 2

    \frac{4}{1/-1} so -4?
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  5. #5
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    Hello, XIII13Thirteen!

    \lim_{x\to2}\frac{x^2-4}{\ln(x^2-3)}
    Be careful . . .

    The derivative of x^2-4 is: . 2x

    The derivative of \ln(x^2-3) is: . \frac{2x}{x^2-3}

    So we have: . \lim_{x\to2}\left(\frac{2x}{\frac{2x}{x^2-3}}\right)  \;=\;\lim_{x\to2}(x^2-3) \;=\;\boxed{1}

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  6. #6
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    Quote Originally Posted by Soroban View Post
    Hello, XIII13Thirteen!

    Be careful . . .

    Thanks, I forgot about the chain rule in the derivative of the denominator
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