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Simple Simple question

  1. Sep 19, 2007 #1


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    1. The problem statement, all variables and given/known data

    lim x -> negative infinity (2x - 4) / rad (3x^2-5)

    2. Relevant equations

    no equations

    3. The attempt at a solution

    I cannot figure out how to rationalize the denominator. I tried rad (3x^2+5) but i am not sure what to do after that. Please solve using ALGEABRA
  2. jcsd
  3. Sep 19, 2007 #2


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    Homework Helper

    You won't rationalize the denominator by using rad(3x^2 *+* 5 ), because that will just give you a difference of two squares under the root sign.

    In fact, you don't need to rationalize this at all. Instead, multiply numerator and denominator by (1/x):

    (2x - 4) · (1/x)

    (1/x) · sqrt(3x^2 - 5) ,

    change the (1/x) in the denominator to the radical sqrt[1/(x^2)] and multiply numerator and denominator through. Now there is a *catch* for the limit x-> negative infinity (or any negative value, really); since we are following *negative* values of x, the square root of (x^2), for instance, is going to be -x [because the square root operation gives a positive value, but our x's are negative]. So, to evaluate *this* limit, we must use (1/x) = -sqrt[1/(x^2)] :

    (2x - 4) · (1/x)
    _________________________ .

    -sqrt[1/(x^2)] · sqrt(3x^2 - 5)

    (2 - [4/x])

    -sqrt(3 - [5/{x^2}]) .

    From here, we can now just use the Limit Law,

    lim x-> plus or minus inf. [1/(x^p)] = 0 , for p positive, to obtain

    (2 - 0)/[ -sqrt(3 - 0) ] = -2/sqrt(3), or after rationalizing, -[2 sqrt(3)]/3 .

    If you were evaluating the limit for x-> plus infinity, that minus sign for the square root wouldn't be needed and the limit would be +[2 sqrt(3)]/3 .

    This illustrates an interesting behavior of rational functions with even roots in them. By now, you are probably used to rational functions of polynomials having just one horizontal asymptote (limit at infinity). When you have a numerator or denominator with a square root of an expression (or fourth root, etc.), however, because of the sign change for negative x, you often end up with *two* horizontal asymptotes, one for each sign of infinity.
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