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Math Help - Finding the limit as x tends to infinity

  1. #1
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    Finding the limit as x tends to infinity

    The question:
    Determine the limiting behaviour of \frac{ln(x^3+1)}{ln(x^2+1)} as x -> infinity.

    My attempt:
    I noticed that both the numerator and denominator approach infinity, making the limit indeterminate. Thus, I tried using L'Hopitals rule. However, in doing so, I can't seem to get an answer even after several differentiations. It seems to constantly become 0/0. Am I attempting this the right way?

    Thanks!
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  2. #2
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    Using L'Hospital...

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

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

     = \lim_{x \to \infty}\frac{3x^4 + 3x^2}{2x^4 + 2x}.

    This is still \frac{\infty}{\infty} so use L'Hospital again...

    \lim_{x \to \infty}\frac{3x^4 + 3x^2}{2x^4 + 2x} = \lim_{x \to \infty}\frac{12x^3 + 6x}{8x^3 + 2}.

    This is still \frac{\infty}{\infty} so use L'Hospital again...

    \lim_{x \to \infty}\frac{12x^3 + 6x}{8x^3 + 2} = \lim_{x\to\infty}\frac{36x^2 + 6}{24x^2}.

    This is still \frac{\infty}{\infty} so use L'Hospital again

    \lim_{x \to \infty}\frac{36x^2 + 6}{24x^2} = \lim_{x \to \infty}\frac{72x}{48x}

     = \lim_{x \to \infty}\frac{3}{2}

     = \frac{3}{2}.
    Last edited by Prove It; September 9th 2010 at 02:02 AM.
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  3. #3
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    Ahh darn, looks like I made an error right at the start. Thanks!

    Also, you can stop after the first application of L'Hopital, since you can divide by the largest powers to get 3/2. (that's a shortcut we learnt at uni)
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    Quote Originally Posted by Glitch View Post
    Ahh darn, looks like I made an error right at the start. Thanks!

    Also, you can stop after the first application of L'Hopital, since you can divide by the largest powers to get 3/2. (that's a shortcut we learnt at uni)
    is 3/2 a typo? denominator has higher degree polynomial so limit is 0.
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  5. #5
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    Prove It had the right answer (3/2), then he changed it for some reason.
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    \lim_{x \to \infty}\frac{\ln{(x^3 + 1)}}{\ln{(x^2 + 1)}} = \lim_{x \to \infty}\frac{\frac{3x^2}{x^3 + 1}}{\frac{2x}{x^2 + 1}} =  \frac{3x^2}{x^3 + 1}.\frac{x^2 + 1}{2x} =  \frac{3x^2(x^2 + 1)}{2x(x^3 + 1)} = \frac{3x(x^2 + 1)}{2(x^3 + 1)} = \frac{3x^3 + 3x}{2x^3 + 2} = \frac{3}{2}

    Unless I'm mistaken. (the textbook states 3/2 as the answer)
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  7. #7
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    Gah hang on, I'll fix it. It's well known my algebra is shocking...
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  8. #8
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    I assure you mine is worse. :P
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  9. #9
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    Quote Originally Posted by Glitch View Post
    Prove It had the right answer (3/2), then he changed it for some reason.
    Ah, didn't bother checking the algebra, just saw that at the end of the first application of L'Hopital's rule Prove It had a cubic over a quartic (now corrected).
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