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Finding a limit help.

  1. Dec 3, 2009 #1
    1. The problem statement, all variables and given/known data

    limit as x->1/3 of (2-6x)^2/(3x-1)(9x^2-1)

    2. Relevant equations

    3. The attempt at a solution

    substituting gets 0/0, tried expanding but it doesnt work either, dunno what else is there left to do...
  2. jcsd
  3. Dec 3, 2009 #2
    can you use l'hopital's rule?

    Edit: l'hopital's rule won't necessarily work for this one forget it

    try simplifying the numerator and then factoring it again while recognizing there is a difference of squares in the denominator

    my explanation might be vague but i don't want to give it away
    Last edited: Dec 3, 2009
  4. Dec 3, 2009 #3
    simplifying the numerator gives 36x^2-24x+4
    factor it to (12x+4)(3x+1)
    factor it again to 4(3x+1)(3x+1)

    can i cancel (3x+1) and (3x-1) ?
    well i tried canceling it anyway and then substituted 1/3 again now i get 8/0 so its +infinity i guess.
  5. Dec 3, 2009 #4
    actually i factored out the numerator a different way
    i took out the 4 from the beginning and got 4(9x^2 - 6x + 1) then got (3x-1)(3x-1)
    you neglected the fact that it -24x, not +24x when you factored it
    for the denominator: (3x-1)(9x^2 - 1)
    recognize the difference of squares in the denominator?
    it will become(3x-1)(3x-1)(3x+1), so cancel and evaluate

    just curious do you know about l'hopital's rule?
    if you do, try to evaluate the limit using that rule and you see that the answer is different:cool:

    just another example of when l'hopital's rule fails if you're interested in knowing that
  6. Dec 3, 2009 #5


    Staff: Mentor

    No, this should be 4(3x - 1)(3x -1).
    It would have been simpler to recognize that (2 - 6x)^2 = (6x - 2)^2 = 4(3x - 1)^2.
    Can you cancel 3 + 1 and 3 - 1?
  7. Dec 3, 2009 #6
    yea your right... im not too good at algebra anyway so it takes time till i realize things like that
  8. Dec 3, 2009 #7
    yea i got 4(9x^2 - 6x + 1) the first time as well but didnt know what to do with it further.. im bad at factoring
    i heard about l'hopital's rule but never learned how to use it.. my prof. didnt teach us cuz he said sometimes it doesnt give the right answer so why bother
  9. Dec 3, 2009 #8
    did you see my post above? kinda pointed that out and helped you out a little more
  10. Dec 3, 2009 #9
    yea man thanks!
  11. Dec 3, 2009 #10


    Staff: Mentor

    I hope you realized that when I asked "can you cancel 3 + 1 and 3 - 1" you understood me to mean that you can't do that. And you can't cancel 3x + 1 and 3x - 1.

    So what did you get as your final expression before you took the limit?
  12. Dec 3, 2009 #11
    hey mark44, i was wondering if you could evaluate the limit using l'hopital's rule and see if I was right or not about it not working for this limit it'll probably only take you 5 seconds

    Thanks a lot.
  13. Dec 3, 2009 #12
    yea i got it thanks!

    i looked at the question and did it all over again and i certainly missed that i should have done 4(3x - 1)(3x -1) instead of 4(3x +1)(3x +1)
    and for the final expression i got:


    which gives 2 after substituting.
  14. Dec 3, 2009 #13


    Staff: Mentor

    That's the right answer, but you can't just substitute 1/3 in because the numerator will be 0 and so will the denominator. That's where the limit comes in. As long as x is only close to 1/3, but not equal to it, then (3x - 1)^2 over itself will be close to 1, and the rest will be close to 4/2 = 2.
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