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Math Help - Limits Help

  1. #1
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    Limits Help

    Hi there,
    just pondering this equation: lim (3x^3(-6x^2)+5x+35)/((x^4)-5x^3-27) when x->∞

    what is the best way to figure this out? I tried to divide everything by the highest factor I would find in the denominator, like my textbook said, but that just got messy and confusing. Any ideas?
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  2. #2
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    Re: Limits Help

    Is the limit actually:

    \lim_{x\to\infty}\frac{3x^3-6x^2+5x+35}{x^4-5x^3-27} ?

    If so, you might just observe that the limit is zero since the degree in the denominator is greater than that of the numerator.
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  3. #3
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    Re: Limits Help

    oh, ok then, that means that I did get it right (I came out with 0/1), but what confused me was that wolfram indicated it was -infinity. Glad I seem to have done the right thing then!!
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  4. #4
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    Re: Limits Help

    I don't see how you could have gotten \displaystyle \begin{align*} \frac{0}{1} \end{align*}. The reason the limit is \displaystyle \begin{align*} 0 \end{align*} is because the denominator goes to \displaystyle \begin{align*} \infty \end{align*} while the numerator does not. Dividing a number by a much larger number gives you something very close to 0.
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: Limits Help

    If you typed it into the text box at wolfram as you did in your first post, then you would get -infinity. Do you see why?
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  6. #6
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    Re: Limits Help

    ahh, it was all those unnecessary brackets I put in... silly me
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  7. #7
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    Re: Limits Help

    in that particular wolfram working, it divides each term by a different power... why is this? I thought we would only divide by x^4??Limits Help-wolfram.png
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  8. #8
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    Re: Limits Help

    They DID divide top and bottom by \displaystyle \begin{align*} x^4 \end{align*}...
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  9. #9
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    Re: Limits Help

    so what about where they divided by x^3, x^2?? Sorry i know I sound like a complete idiot, but these things take time to sink in
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  10. #10
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    Re: Limits Help

    \displaystyle \begin{align*} \frac{3x^3 - 6x^2 + 5x + 35}{x^4 - 5x^3 - 27} &= \frac{\frac{3x^3 - 6x^2 + 5x + 35}{x^4}}{\frac{x^4 - 5x^3 - 27}{x^4}} \\ &= \frac{\frac{3x^3}{x^4} - \frac{6x^2}{x^4} + \frac{5x}{x^4} + \frac{35}{x^4}}{\frac{x^4}{x^4} - \frac{5x^3}{x^4} - \frac{27}{x^4}} \\ &= \frac{\frac{3}{x} - \frac{6}{x^2} + \frac{5}{x^3} + \frac{35}{x^4}}{1 - \frac{5}{x} - \frac{27}{x^4}} \end{align*}
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  11. #11
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    Re: Limits Help

    Oh, now I get it!! Thank you so much!
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  12. #12
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    Re: Limits Help

    Quote Originally Posted by NettieL View Post
    Oh, now I get it!! Thank you so much!
    Also a good thing to remember when solving limits that tend to infinity is that you can take the highest power in the numerator over the highest power in the denominator and calculate the limit of that. In your case it would be 3x^3/x^4 which would simplify to 3/x and the limit of 3/x as x tends to -infinity is obviously 0.
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