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Math Help - please very urgent

  1. #1
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    please very urgent

    1> lim ((3x^3)-2x) / ((x^2)-1)
    x->infinite



    2> lim 1/ |x-3|
    x->3-
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by bobby77
    1> lim ((3x^3)-2x) / ((x^2)-1)
    x->infinite
    \lim_{x \to \infty} \frac{3x^3-2x}{x^2-1}=\lim_{x \to \infty} \frac{3x-2/x}{1-1/x^2}<br />

    So as x\to \infty the top \to \infty and the
    bottom \to 1, so:

    <br />
\lim_{x \to \infty} \frac{3x^3-2x}{x^2-1}=\infty<br />

    which is short hand for it diverges.

    RonL
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  3. #3
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    \lim_{x\to\infty}\frac{3x^3-2x}{x^2-1}

    Look at the highest power of the numerator versus the denominator. The numerator has a degree of 3 while the denominator has a degree of two. You should see that as as grows larger and larger, the top far outgrows the bottom and this limit is infinity.

    \lim_{x\to{3-}}\frac{1}{|x-3|}

    You should first realize that this is a form of the rational graph \frac{1}{x} shifted to the right 3 units, and all negative values are flipped about the x-axis. Here's how I would do this problem. Test the value -2.9. How big is that number? Is it positive or negative? Now test -2.99, then -2.9999. See the trend?
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