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Math Help - Limits involving square roots

  1. #1
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    Limits involving square roots

    Hello. I've never been great at math (or spent much time studying), but, since my new major requires quite a few math classes, I have decided to put in as much time as I can to develop my skills. I need help finding a limit for a problem involving square roots. The problem is probably very simple, but I am not sure exactly how to work with the square roots/squares in the problem. I have tried as many approaches as I can think of, and I still feel like I'm getting nowhere.

    Could anyone walk me through the steps in solving this? That would be EXTREMELY helpful. The problem:


    lim t->2

    sqrt( (t+4)(t-2)^4 )
    -------------------------
    (3t - 6)^2


    Sorry, I have no idea how to format that properly. Thanks for any help in advance!
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  2. #2
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    Re: Limits involving square roots

    Quote Originally Posted by nikdsm2012 View Post
    Hello. I've never been great at math (or spent much time studying), but, since my new major requires quite a few math classes, I have decided to put in as much time as I can to develop my skills. The problem:

    lim t->2

    sqrt( (t+4)(t-2)^4 )
    -------------------------
    (3t - 6)^2N
    No walk through is available here. We don't do that.

    But \(3t-6)^2= 9(t-2)^2 so \frac{\sqrt{(t+4)(t-2)^4}}{(3t-6)^2}=\frac{\sqrt{(t+4)}(t-2)^2}{9(t-2)^2}=\frac{\sqrt{(t+4)}}{9}
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  3. #3
    MHF Contributor ebaines's Avatar
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    Re: Limits involving square roots

    If this is what you mean:

     \lim _{t \to 2} \frac {\sqrt{(t+4)(t-2)^4}}{(3t-6)^2}

    You can start by pulling the (t-2)^4 term out of the square root, then simplify:

     \frac {\sqrt{(t+4)(t-2)^4}}{(3t-6)^2} = \frac {\sqrt{t+4}(t-2)^2}{3^2(t-2)^2} = \frac {\sqrt{t+4}} 9

     \lim _{t  \to 2} \frac {\sqrt{t+4}} 9 = \frac {\sqrt 6 } 9
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