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Homework Help: LIMIT question needs your attention

  1. Sep 15, 2011 #1
    Limit x-->0- [[sqrt(88*x^2)] / x]
    i think the answer of this question should be sqrt(88) but according to my instructor its -sqrt(88)
    so please solve this one for me. only the left hand limit .
     
  2. jcsd
  3. Sep 15, 2011 #2

    LCKurtz

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    Remember that sqrt(x2)= |x| = -x if x < 0.
     
  4. Sep 15, 2011 #3
    ok...so the x in numerator will be positive and the one in the denominator negative.
    so now the answer will be -sqrt(88)
     
  5. Sep 15, 2011 #4

    LCKurtz

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    Not quite. The x in the numerator is equal to the x in the denominator; it's not that one of them is negative and the other positive. The point is that the fraction |x|/x = (-x)/x. Still, x itself is negative as x → 0-.
     
  6. Sep 15, 2011 #5
    i have one more question for you
    cant we cut the two 'x' in the numerator and denominator without putting values?
    if we do so then we wont get the - sign in the answer
     
  7. Sep 15, 2011 #6

    LCKurtz

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    You can't cancel the x values in |x|/x without distinguishing the cases whether x > 0 or x < 0.
     
  8. Sep 15, 2011 #7
    sorry.i get your point now. but your last point has confused them in another problem:
    lim x->5 [3x-15/(sqrt{x^2 + 25 - 10x}]
    how will u solve this question for the left and right hand limits separately?
     
  9. Sep 15, 2011 #8

    LCKurtz

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    Given our previous example, what do you get for your denominator [itex]\sqrt{(x-5)^2}[/itex] ? It is the same idea.
     
  10. Sep 15, 2011 #9
    here is what i understand:
    for right hand limit : 3(x-5)/|x-5|
    and since both numerator and denominator are positive we get 3

    for left handed : 3(x-5)/|x-5|
    in this case the numerator becomes negative but the denominator stays positive . so we get -3
    ryt?
    and overall the limit of function doesnt exist at 3
     
  11. Sep 15, 2011 #10

    LCKurtz

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    :smile: Yes that's right. Sorry for the delay getting back to you but the rest of my life intervened.
     
  12. Sep 16, 2011 #11
    thanks for your help...you cleared my concept
     
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