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Find this limit

  1. Jan 4, 2008 #1
    hi,
    find this limit
     

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  2. jcsd
  3. Jan 4, 2008 #2
  4. Jan 4, 2008 #3

    HallsofIvy

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    Yes, the fact that [itex]x\sqrt{A}= \sqrt{x^2A}[/itex] (and the fact that square root is continuous wherever it is defined) makes the problem simple.

    In any case, m s a, this is clearly school work which should have been posted in the "Homework and Coursework" section so I am moving it there.

    Also, as you should already know, you MUST show what work you have already done and what thoughts you had about it so we will KNOW what help you need.
     
  5. Jan 4, 2008 #4

    CompuChip

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    Hi.

    I found it.

    Can I keep it?
     
  6. Jan 4, 2008 #5

    stewartcs

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    Becareful, this is only true when [tex]x \geq 0[/tex].
     
  7. Jan 4, 2008 #6
    Thank you all tried to find the end

    but the fee is quite different
    These endeavours
    Where is the error??
     

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  8. Jan 4, 2008 #7

    stewartcs

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    I don't understand what you just wrote. :confused:

    Can you properly state your question please?

    CS
     
  9. Jan 4, 2008 #8
    cont.
     

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  10. Jan 4, 2008 #9
    chart shows that the end does not exist
     
  11. Jan 4, 2008 #10
    This fee Key
    clear that the end does not exist
    is not it?
     

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  12. Jan 4, 2008 #11

    HallsofIvy

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    Yes,. the problem is what stewartcs pointed out. If x> 0, then [itex]x\sqrt{4+ 1/x^2}= \sqrt{4x^2+ 1}[/itex] and that has limit 1. But is x< 0, then [itex]x\sqrt{4+ 1/x^2}= -\sqrt{4x^2+ 1}[/itex] and that has limit -1. The "limit from above" is 1 and the "limit from below" is -1. Since those are different, the limit itself does not exist.
     
    Last edited: Jan 4, 2008
  13. Jan 4, 2008 #12
    Thank you again
    I will return to this subject
     
  14. Jan 4, 2008 #13

    stewartcs

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    Correct, the limit does not exist.

    CS
     
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