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2 radicals in a limit?

  1. Sep 21, 2013 #1
    1. The problem statement, all variables and given/known data
    limx->4 (sqrt(5-x)-1) / (2-sqrt(x))

    NOT ALLOWED TO USE L'HOSPITALS.


    2. Relevant equations



    3. The attempt at a solution


    I tried using conjugate of both top and bottom and I couldn't get it to work, but maybe I've done it wrong?.. I'm not allowed to use L'Hospital's rules so how can I solve this?
     
  2. jcsd
  3. Sep 21, 2013 #2

    haruspex

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    can you think of something you can multiply top and bottom by to get rid of the radical below the line? Does that give you any further ideas?
     
  4. Sep 21, 2013 #3
    just by x? Because if I multiply by (2+sqrt(x)) it just got long and tedius, thought there was an easier way.
     
  5. Sep 21, 2013 #4

    SteamKing

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    We can't tell if you made a mistake in using the conjugate unless you post your work.
     
  6. Sep 21, 2013 #5
    When I multiplied by the conjugate I ended up with 4-x on the denominator. Subbing in 4 would make the limit impossible.
     
  7. Sep 21, 2013 #6

    HallsofIvy

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    Multiplying both numerator and denominator by [itex]2+\sqrt{4}[/itex] will give an "[itex]x- 4[/itex]" in the denominator but multiplying both numerator and denominator by [itex]\sqrt{5- x}+ 1[/itex] gives "[itex](5- x)- 1= -(x- 4)[/itex]" in the numerator. What can you do now?
     
  8. Sep 21, 2013 #7
    The only thing I see you can do with that is square it I guess?I'm not sure... Subbing in 4 will just make it 0.
     
  9. Sep 21, 2013 #8

    eumyang

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    No. I think what HallsofIvy is trying to tell you is that you have to multiply by the conjugate twice. See my post in this thread.
     
  10. Sep 22, 2013 #9
    By rationalize do you mean multiply the conjugate? I don't see how you can rationalize (multiply top and bottom by the root), but you say you only do it to the top and then only do it to the bottom. I don't think you can do that. Also if I multiply the top conjugate and then bottom conjugate it gets REALLY messy and it can't be right.

    (This is all after reading your post)
     
  11. Sep 22, 2013 #10

    haruspex

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    You are not posting your working, so I have to guess. It sounds like you went through the full multiplication top and bottom each time. Don't do that. Multiply top and bottom by that expression which removes the radical from the denominator, but in the numerator just leave it as a product of two radical expressions. don't multiply that out. Now do the same for the radical expression which is the original numerator - multiply top and bottom by its radical 'conjugate'. You should now have an expression which is a product of two terms in the numerator (one being the conjugate of the original denominator, and the other not involving a radical) and similarly two terms in the denominator.
    Now look at the two non-radical terms. Do you see any cancellation?
     
  12. Sep 22, 2013 #11
    If you say don't multiply the 2 radical expressions out, am I leaving the (sqrt(5-x)-1)(2+sqrt(x)) in the top, then multiplying the top and bottom by sqrt(5-x)+1? So in the end I'd be multiplying 3 different expressions in the top:
    (sqrt(5-x)-1)(2+sqrt(x))(sqrt(5-x)+1).

    Here is what I get after multiplying by the conjugate of 2-sqrt(x). Then I plugged in the next conjugate and it just seems like it would get too messy.
    http://i1301.photobucket.com/albums/ag115/phrox1/20130922_123901_zps9183e7f8.jpg [Broken]



    I have to be missing something here, it's not clicking in my head for some reason.
     
    Last edited by a moderator: May 6, 2017
  13. Sep 22, 2013 #12
    nevermind I understand now, thanks
     
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