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Math Help - Finding Limit as X approaches INFINITY

  1. #1
    Senior Member topher0805's Avatar
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    Finding Limit as X approaches INFINITY

    Find the Limit:




    I multiplied the numerator and denominator by the denominator, then started factoring things out and I ended up getting 1/64.

    WebAssign is telling me this is wrong, but I can't figure out why. Any help?
    Last edited by topher0805; February 6th 2008 at 08:32 PM.
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    Quote Originally Posted by topher0805 View Post
    Find the Limit:




    I multiplied the numerator and denominator by the denominator, then started factoring things out and I ended up getting 1/64.

    WebAssign is telling me this is wrong, but I can't figure out why. Any help?
    Things can get tricky with the -\infty. To avoid trouble, make the substitution t = -x. Then you want


    \lim_{t \rightarrow +\infty} (-t + \sqrt{t^2 - 2t}) = \lim_{t \rightarrow +\infty} (\sqrt{t^2 - 2t} - t).


    Note that:


    \displaystyle \frac{(\sqrt{t^2 - 2t} - t)(\sqrt{t^2 - 2t} + t)}{\sqrt{t^2 - 2t} + t} = \frac{(t^2 - 2t) - t^2}{\sqrt{t^2 - 2t} + t} = \frac{-2t}{\sqrt{t^2 - 2t} + t} = \frac{-2}{\sqrt{1 - \frac{2}{t}} + 1}.


    \lim_{t \rightarrow +\infty} \frac{-2}{\sqrt{1 - \frac{2}{t}} + 1} = \frac{-2}{1 + 1} = -1.
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    Senior Member topher0805's Avatar
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    Well, I did need help with that one too but the one I was actually describing is now in my original post.
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    Quote Originally Posted by topher0805 View Post
    Well, I did need help with that one too but the one I was actually describing is now in my original post.
    You can't just do an edit and completely change the question that's been replied to! It's poor form. Good job I quoted the original original post.

    New question - new post.
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  5. #5
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    Quote Originally Posted by topher0805 View Post
    Find the Limit:




    I multiplied the numerator and denominator by the denominator, then started factoring things out and I ended up getting 1/64.

    WebAssign is telling me this is wrong, but I can't figure out why. Any help?
    Divide top and bottom by x and then take the limit:

    \lim_{x \rightarrow \infty}\frac{x + 2}{\sqrt{64x^2 + 1}} = \lim_{x \rightarrow \infty}\frac{1 + \frac{2}{x}}{\sqrt{64 + \frac{1}{x^2}}} = \frac{1 + 0}{\sqrt{64 + 0}} = \frac{1}{8}
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