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Math Help - Find the horizontal asymptote

  1. #1
    DBA
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    Find the horizontal asymptote

    f(x)= [sqrt(2x^2 +1)] / [3x-5]

    for x-> infinity, we have x>0

    I need to divide numerator and denominator by the highest power of the denominator, which is x in this case.

    lim x-> infinity [sqrt(2x^2 +1)] / [3x-5]

    The solution in the book has the following step

    lim x-> infinity [sqrt(2 +1/x^2)] / [3-5/x]

    I do not understand how they get to this numerator, when dividing by x?


    The book gives also:
    For x>0 --> sqrt(x^2) = x
    For x<0 --> sqrt(x^2) = |x| = -x

    Can someone explain that to me. What steps are in between.
    Thanks
    Last edited by DBA; October 26th 2009 at 01:45 PM. Reason: additional information
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  2. #2
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    As you noted, you want to divide by the highest power in the denominator which is x. However, "x" isn't the only way to represent "x". Is not \sqrt{x^{2}} (for x greater than 0) also a means of denoting "x"?
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  3. #3
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    Quote Originally Posted by DBA View Post
    The solution in the book has the following step

    lim x-> infinity [sqrt(2 +1/x^2)] / [3-5/x]

    I do not understand how they get to this numerator, when dividing by x?
    You are missing an important term, but this is almost correct.

    What they did is factor out an x^2 from x^2+1, which becomes x^2(1+[1\x^2]). Now you can apply the square root to the x^2 term and demonstrate that the powers of the numerator and denominator are equal. Also notice that 1/x^2 tends to 0 as x goes to infinity.
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