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Math Help - theorem of L'Hospital

  1. #1
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    theorem of L'Hospital

    Using the theorem of L'Hospital, calculate:

    \lim_{n\to\infty} \frac{ln(x^2+x+1)}{ln(x^3+2x+1)}


    answer:
    \frac{2}{3}
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  2. #2
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    Quote Originally Posted by Apprentice123 View Post
    Using the theorem of L'Hospital, calculate:

    \lim_{n\to\infty} \frac{ln(x^2+x+1)}{ln(x^3+2x+1)}


    answer:
    \frac{2}{3}
    I presume you mean
    \lim_{x \to \infty} \frac{ln(x^2 + x + 1)}{ln(x^3 + 2x + 1)}

    This form is of the type \infty / \infty so we can use L'Hopital.
    \lim_{x \to \infty} \frac{ln(x^2 + x + 1)}{ln(x^3 + 2x + 1)}

    = \lim_{x \to \infty} \frac{\frac{1}{x^2 + x + 1} \cdot (2x + 1)}{\frac{1}{x^3 + 2x + 1} \cdot (3x^2 + 2)}

    = \lim_{x \to \infty} \frac{(x^3 + 2x + 1)(2x + 1)}{(x^2 + x + 1)(3x^2 + 2)}

    This is still indeterminate, but we can now multiply out the numerator and denominator and are left with a polynomial divided by a polynomial. I'll leave the rest to you.

    -Dan
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  3. #3
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    Thanks for helping
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