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Homework Help: Pls help:math question

  1. Dec 29, 2005 #1
    lim x to 9 [(x^2]-81)/[3-(3^1/2)]??

    thanx.......
     
  2. jcsd
  3. Dec 29, 2005 #2

    siddharth

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    Can you show what you have done till now with this problem?
     
  4. Dec 29, 2005 #3
    i have tried to do this question n got the answer of -108.is it correct??
     
  5. Dec 29, 2005 #4

    siddharth

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    No, it is not right.
    If you show exactly how you got your answer of -108 (ie, post all the working and calculations you did) it will be easier for us to help you and show where you went wrong.
     
  6. Dec 29, 2005 #5

    TD

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    That answer isn't correct, unless you stated the wrong problem. I *think* you may mean

    [tex]\mathop {\lim }\limits_{x \to 9} \frac{{x^2 - 81}}{{3 - \sqrt x }}[/tex]

    In that case, your answer is correct :smile:
     
  7. Dec 29, 2005 #6
    ya,that is my question.thank you very much.....
     
  8. Dec 29, 2005 #7

    TD

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    No problem, but be careful when "writing math" ;)
     
  9. Dec 29, 2005 #8
    yeah, try using latex... just click on the equation TD wrote and look at the code, its worth learning.
     
  10. Dec 30, 2005 #9

    HallsofIvy

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    Okay, the problem is
    [tex]\mathop {\lim }\limits_{x \to 9} \frac{{x^2 - 81}}{{3 - \sqrt x }}[/tex]

    do you now see how to get the limit?

    The problem is that if you just replace x by 9, both numerator and denominator are 0. (If the denominator were not 0, just calculate the value at x= 9. If the denominator is 0 but the numerator not, there is no limit.)

    But since both become 0 at x= 9, that means we can factor! x2- 81 is a "difference of squares" so x2- 81= (x- 9)(x+ 9) and then we can treat x- 9 as a "difference of squares also: [itex]x- 9= (\sqrt(x))^2- 3^2= (\sqrt{x}- 3)(\sqrt{x}+ 3)[/itex]
    Can you finish it from there?
     
  11. Dec 30, 2005 #10
    well i think he did even before posting the question, because he got a right answer, and just asked if its right.
     
  12. Dec 30, 2005 #11

    HallsofIvy

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    Yeah, I just reread the original post and that is implied.
     
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