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Math Help - Negative limits

  1. #1
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    Negative limits

    Lim (x>-infinity) (-3x + 1)/(x^2 + x)^(1/2).

    I think it should be 3, but I don't know how to show it.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by unluckykc View Post
    Lim (x>-infinity) (-3x + 1)/(x^2 + x)^(1/2).

    I think it should be 3, but I don't know how to show it.
    as x \to - \infty, the 1 in the top and the x in the bottom do not really make a difference, so we can get rid of them.

    thus, \lim_{x \to - \infty} \frac {-3x + 1}{\sqrt{x^2 + x}} = \lim_{x \to - \infty} \frac {-3x}{\sqrt{x^2}} = \lim_{x \to - \infty} \frac {-3x}{|x|} = \left \{ \begin{array}{lr} -3 & \mbox{ as } x \to \infty \\ 3 & \mbox{ as } x \to -\infty \end{array}\right.
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  3. #3
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    Since \sqrt{x^{2}}=|x| we can divide by -x and \sqrt{x^{2}} where we need to.

    \lim_{x\rightarrow{-\infty}}\frac{\frac{-3x}{-x}+\frac{1}{-x}}{\sqrt{1+\frac{1}{x}}}

    Now, see the limit?.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    and again, we could split the original into two pieces.


    \frac {-3x + 1}{\sqrt{x^2 + x}} = \frac {-3x }{\sqrt{x^2 + x}} + \frac 1{\sqrt{x^2 + x}}

    now the second fraction goes to zero as x \to - \infty so we can forget about it. the first fraction we can simplify into \frac {-3x}{|x|} as x \to - \infty and the conclusion follows
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