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March 30, 2017

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Evaluate the limit. as x approaches infinity, lim sqrt(x^2+6x+3)-x

  • Calculus - ,

    sqrt(x^2+6x+3) =

    x sqrt(1 + 6/x + 3/x^2)

    Using the series expansion:

    (1+y)^p = 1 + p y + p (p-1)/2 y^2 + ...

    for p = 1/2 gives:

    sqrt(1 + 6/x + 3/x^2) = 1 + 3/x +

    O(1/x^2)

    Where the O(1/x^2) means that there exists an R and a constant c such that for x larger than R, the difference between the square root and (1+3/x) becomes less than c/x^2.

    We then have:

    x sqrt(1 + 6/x + 3/x^2) - x =

    3 + O(1/x)

    This means that the limit for x to infinity is 3.

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