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Factoring A difference of fifths

  1. Sep 12, 2011 #1
    One of the homework questions i did tonight was (x^5-32)/(x-2), where i had to find the limit when x approaches two. In order to do this i looked up the formula to use for a difference of fifths and was able to solve the question by factoring out the denominator and putting two in for x, giving me an answer of 80.

    I am wondering if there is a much simpler way of solving this which i foolishly missed as we are definitely not expected to know the formula for a difference of fifths and i'm not sure it is easily derived. If anyone can spot one or argue that there isn't one that would be great! :)
  2. jcsd
  3. Sep 12, 2011 #2


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    L'hôpital's rule is probably the easiest.

    Doing the polynomial division is a close second, I think.
  4. Sep 12, 2011 #3
    having not learned derivatives yet i suppose the method i indicated is the most suitable then?
  5. Sep 13, 2011 #4


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    It is generally true that [itex]x^n- y^n= (x- y)(x^{n-1}+ x^{n-2}y+ \cdot\cdot\cdot+ xy^{n-2}+ y^{n-1})[/itex].

    That seems the best method to me. You should not have to "look it up".
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