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Thread: lim x->∞/-∞?

  1. #1
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    lim x->∞/-∞?

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

    Question: Have you graphed the function?

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

    What does the sign of $\displaystyle f'(x)$ when $\displaystyle x<0$ tell you?
    Last edited by Plato; Nov 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.


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

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

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


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

    . . $\displaystyle \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 $\displaystyle x\!:$

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

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


    Therefore: .$\displaystyle \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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    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: .$\displaystyle \frac{\sqrt{x^2-2x} - x}{1}$


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

    . . $\displaystyle \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 $\displaystyle x\!:$

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

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


    Therefore: .$\displaystyle \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 $\displaystyle x\to\infty$, but not for the limit as $\displaystyle x\to-\infty$.
    ..
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