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Math Help - Question about Limits involving infinity

  1. #1
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    Question about Limits involving infinity

    The problem is: evaluate the limit as x approaches infinity of [x-sqrt(x)]. What I did was multiply that by [x+sqrt(x)]/[x+sqrt(x)] to get (x^2-x)/[x+sqrt(x)], i then factored an x out of the top, getting [x(x-1)]/sqrt(x). Then, since the degree of the numerator is higher, I said that the limit is infinity. I know that I got the right answer, but did I do the problem in the correct way?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Is...

    x - \sqrt{x} = \sqrt{x} (\sqrt{x}-1) = \frac{\sqrt{x}}{\sqrt{x}+1} (x-1) (1)

    ... and the You have to valuate the limit for x \rightarrow \infty of (1)...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by jem163 View Post
    The problem is: evaluate the limit as x approaches infinity of [x-sqrt(x)]. What I did was multiply that by [x+sqrt(x)]/[x+sqrt(x)] to get (x^2-x)/[x+sqrt(x)], i then factored an x out of the top, getting [x(x-1)]/sqrt(x). Then, since the degree of the numerator is higher, I said that the limit is infinity. I know that I got the right answer, but did I do the problem in the correct way?
    Your calculations are valid, although most of the work is unnecessary to be honest. You can consider the original expression as

    \frac{x-x^{1/2}}{1}

    Clearly already the numerator is higher degree than the denominator, and the x term is the term of highest degree in the numerator, so it defines the end-behavior.
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