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Math Help - lim x->∞/-∞?

  1. #1
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    lim x->∞/-∞?

    Why is it that lim x->∞ \sqrt {x^2-2x}-x = 1 while lim x->-∞ \sqrt {x^2-2x}-x =∞? They both simplify to the same function but one gives a numerical value while the other one doesn't.
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  2. #2
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    Re: lim x->∞/-∞?

    Quote Originally Posted by Barthayn View Post
    Why is it that lim x->∞ \sqrt {x^2-2x}-x = 1 while lim x->-∞ \sqrt {x^2-2x}-x =∞? They both simplify to the same function but one gives a numerical value while the other one doesn't.
    First a correction: let f(x)=\sqrt{x^2-2x}-x then
    \lim _{x \to \infty } f(x) =  - 1.

    Question: Have you graphed the function?

    Do you see that \text{if }x<0\text{ then }f(x)>0~?

    What does the sign of f'(x) when x<0 tell you?
    Last edited by Plato; November 29th 2011 at 12:50 PM.
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  3. #3
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    Re: lim x->∞/-∞?

    Hello, Barthayn!

    I don't understand your problem.
    I get only one answer.


    \text{Why is it that }\lim_{x\to\infty}\left(\sqrt{x^2-2x}-x\right) \:=\: 1

    . . . \text{while }\lim_{x\to -\infty}\left(\sqrt {x^2-2x} -x\right) \:=\:\infty\,?

    We have: . \frac{\sqrt{x^2-2x} - x}{1}


    Multiply by \frac{\sqrt{x^2-2x} + x}{\sqrt{x^2-2x} + x}:\;\;

    . . \frac{\sqrt{x^2-2x} - x}{1} \cdot\frac{\sqrt{x^2-2x} + x}{\sqrt{x^2-2x} + x} \;=\;\frac{(x^2-2x) - x^2}{\sqrt{x^2-2x} + x} \;=\;\frac{-2x}{\sqrt{x^2-2x} + x}


    Divide numerator and denominator by x\!:

    . . \dfrac{\dfrac{-2x}{x}}{\dfrac{\sqrt{x^2-2x}}{x} + \dfrac{x}{x}} \;=\; \dfrac{-2}{\dfrac{\sqrt{x^2-2x}}{\sqrt{x^2}} + 1} \;=\;\dfrac{-2}{\sqrt{\dfrac{x^2-2x}{x^2}} + 1}

    . . =\;\dfrac{-2}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}} + 1}  \;=\;\dfrac{-2}{\sqrt{1 - \dfrac{2}{x}} + 1}


    Therefore: . \lim_{x\to\infty}\dfrac{-2}{\sqrt{1 - \frac{2}{x}} + 1} \;=\;\dfrac{-2}{\sqrt{1+0} + 1} \;=\;\frac{-2}{2} \;=\;-1

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  4. #4
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    Re: lim x->∞/-∞?

    Quote Originally Posted by Soroban View Post
    Hello, Barthayn!

    I don't understand your problem.
    I get only one answer.



    We have: . \frac{\sqrt{x^2-2x} - x}{1}


    Multiply by \frac{\sqrt{x^2-2x} + x}{\sqrt{x^2-2x} + x}:\;\;

    . . \frac{\sqrt{x^2-2x} - x}{1} \cdot\frac{\sqrt{x^2-2x} + x}{\sqrt{x^2-2x} + x} \;=\;\frac{(x^2-2x) - x^2}{\sqrt{x^2-2x} + x} \;=\;\frac{-2x}{\sqrt{x^2-2x} + x}


    Divide numerator and denominator by x\!:

    . . \dfrac{\dfrac{-2x}{x}}{\dfrac{\sqrt{x^2-2x}}{x} + \dfrac{x}{x}} \;=\; \dfrac{-2}{\dfrac{\sqrt{x^2-2x}}{\sqrt{x^2}} + 1} \;=\;\dfrac{-2}{\sqrt{\dfrac{x^2-2x}{x^2}} + 1} Not if x<0, because then \color{red}x\ne\sqrt{x^2} !

    . . =\;\dfrac{-2}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}} + 1}  \;=\;\dfrac{-2}{\sqrt{1 - \dfrac{2}{x}} + 1}


    Therefore: . \lim_{x\to\infty}\dfrac{-2}{\sqrt{1 - \frac{2}{x}} + 1} \;=\;\dfrac{-2}{\sqrt{1+0} + 1} \;=\;\frac{-2}{2} \;=\;-1

    The answer x=1 is correct, of course, for the limit as x\to\infty, but not for the limit as x\to-\infty.
    ..
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