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Limit x->-infinity (x/(z^2+x^2)^0.5) = ?

  1. Jun 22, 2005 #1


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    Hi Ho! :smile:

    Mmmm... I have a problem with this one:

    [itex]\lim_{x\rightarrow-\infty} \frac{x}{\sqrt{z^2+x^2}}[/itex]

    Using a computer graphics tool, I found that the result should be -1 by looking at the generated graph.

    But if I do it by hands, I find 1 as follows:

    [itex]\lim_{x\rightarrow-\infty} \frac{x}{\sqrt{z^2+x^2}} = \lim_{x\rightarrow-\infty} \frac{x \frac{1}{x}}{\sqrt{z^2+x^2} \frac{1}{x}}[/itex]
    [itex]= \lim_{x\rightarrow-\infty} \frac{1}{\sqrt{\frac{z^2+x^2}{x^2}}}[/itex]
    [itex]= \lim_{x\rightarrow-\infty} \frac{1}{\sqrt{1+(\frac{z}{x})^2}}[/itex]
    [itex]= \frac{1}{\sqrt{1+(\frac{z}{-\infty})^2}}[/itex]
    [itex]= \frac{1}{\sqrt{1+0}}[/itex]
    [itex]= \frac{1}{1}[/itex]
    [itex]= 1[/itex]

    Would you please correct my mistake?

    Thank you very much! :biggrin:
    Last edited: Jun 22, 2005
  2. jcsd
  3. Jun 22, 2005 #2
    hello there

    check this out

    [itex]\lim_{x\rightarrow-\infty} \frac{x}{\sqrt{z^2+x^2}}[/itex]
    [itex]=\lim_{n\rightarrow\infty} \frac{-n}{\sqrt{k^2+n^2}}[/itex]
    [itex]= \lim_{n\rightarrow\infty} \frac{-n\frac{1}{n}}{\sqrt{k^2+n^2} \frac{1}{n}}[/itex]
    [itex]= \lim_{n\rightarrow\infty} \frac{-1}{\sqrt{\frac{k^2+n^2}{n^2}}}[/itex]
    [itex]= \lim_{n\rightarrow\infty} \frac{-1}{\sqrt{1+(\frac{k}{n})^2}}[/itex]
    [itex]= \frac{-1}{\sqrt{1+0}}=-1[/itex]

    always try to simplify it to something that looks simple

    take care

  4. Jun 22, 2005 #3


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    I hope you saw your mistake,you shouln't have sent "x" into its square when taking the limit to "-infinity".

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