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Math Help - How to calculate the following limit

  1. #1
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    How to calculate the following limit

    I'm struggling a bit with this one. Can someone please show me the work required to find the result. I know the answer is \frac{7}{16} but I'm not sure how to arrive at that.

    The question is:

    \lim \frac{-7x^2-4x+16}{-16x^2+3x+3} as x \rightarrow -\infty
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  2. #2
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    Quote Originally Posted by softwareguy View Post
    I'm struggling a bit with this one. Can someone please show me the work required to find the result. I know the answer is \frac{7}{16} but I'm not sure how to arrive at that.

    The question is:

    \lim \frac{-7x^2-4x+16}{-16x^2+3x+3} as x \rightarrow -\infty
    You can use L'Hopital's rule here.
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  3. #3
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    Quote Originally Posted by ur5pointos2slo View Post
    You can use L'Hopital's rule here.
    I shouldn't assume you have learned L'Hopital's rule yet. Let me show you.

    lim x-> - infinity (-7x^2-4x+16) / (-16x^2 + 3x + 3)

    Plugging in infinity you get infinity/infinity which is L'Hop material

    so take the derivative of the top and bottom

    lim x-> - infinity (-14x-4) /(-32x+3) again infinity/infinity

    take the derivative of the top and bottom again

    lim x-> - infinity (-14/-32) = 7/16
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  4. #4
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    \lim_{x\to -\infty} \frac{-7x^2-4x+16}{-16x^2+3x+3} = \lim_{x\to -\infty} \frac{-\frac{7x^2}{x^2}-\frac{4x}{x^2}+\frac{16}{x^2}}{-\frac{16x^2}{x^2}+\frac{3x}{x^2}+\frac{3}{x^2}} = \lim_{x\to -\infty} \frac{-7-\frac{4}{x}+\frac{16}{x^2}}{-16+\frac{3}{x}+\frac{3}{x^2}} = \frac{-7-0+0}{-16+0+0} = \frac{-7}{-16} = \frac{7}{16} <br /> <br /> <br />
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