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Evaluate Lim → 0, radicals in the numerator and denominator!

  1. Feb 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Evaluate lim x →0 √(x+1) - √(2x+1)
    -----------------
    √(3x+4) - √(2x+4)

    2. Relevant equations



    3. The attempt at a solution

    First I rationalized the numerator by multiplying everything by √(x+1) + √(2x+1) / √(x+1) + √(2x+1)

    so now i have
    -x
    ----
    ( √(3x+4) - √(2x+4) )(√(x+1) + √(2x+1))


    and I'm stuck from here, the answer is -2 so I'm not sure if I'm even right so far.
    Thanks in advance for the help!
     
  2. jcsd
  3. Feb 18, 2013 #2

    jbunniii

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    Try multiplying the numerator and denominator by ##\sqrt{3x+4}+\sqrt{2x+4}##?
     
  4. Feb 18, 2013 #3

    DrClaude

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    Try L'Hôpital's rule.
     
  5. Feb 18, 2013 #4
    Well we never learned l'hopital's rule yet but after googling it and attempting it i got
    (1/2) - 1
    ----------
    (3/2) - (1/2)
    which gave me -1/4, not the answer.

    Now as for jbunnii's plan. I got up to
    - x ( (√3x+1) + (√2x+4) )
    ------------------------
    (x-3) ((√x+1) + (√2x+1))

    I'm unsure of what to do next.

    Edit: Can the X's cancel out here? so I end up with
    - ( (√3x+1) + (√2x+4) )
    ------------------------
    -3 ((√x+1) + (√2x+1))

    Edit 2 : No that's giving me - 1/2, the inverse of what I need
     
  6. Feb 18, 2013 #5

    DrClaude

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    How can you get -1/4? Anyway, the (3/2) in the denominator is wrong.
     
  7. Feb 18, 2013 #6

    jbunniii

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    How did you end up with ##x-3## in the denominator? I calculate
    $$(\sqrt{3x+4} - \sqrt{2x+4})(\sqrt{3x+4} + \sqrt{2x+4}) = (3x+4) - (2x+4) = x$$
    Now cancel the ##x## in the numerator and denominator and see what is left.
     
  8. Feb 18, 2013 #7

    jbunniii

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    Also, you somehow changed your ##\sqrt{3x+4}## into ##\sqrt{3x+1}##.
     
  9. Feb 18, 2013 #8
    yeah you're right, I got the answer! I just copied the question down wrong. Thanks for the help everyone!
     
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