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Solving limits algebraically

  1. Aug 31, 2006 #1
    I'm trying to solve this limit algebraically

    limit as x approaches 1 (x^.5-x^2)/(1-x^.5).

    I tried multiplying by the conjugate 1+x^.5 but that left the denominator with 1-x and I was unable to cancel that out.

    Can anyone help me out?
     
  2. jcsd
  3. Aug 31, 2006 #2

    Hurkyl

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    Well, can you show what you got when you tried that?
     
  4. Aug 31, 2006 #3
    Sure.

    I got (x^.5+x-x^2-x^(5/2))/(1-x).

    I can't figure out how to factor the top so as to cancel out the 1-x part.
     
    Last edited: Aug 31, 2006
  5. Aug 31, 2006 #4

    Hurkyl

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    Well, you can always do long-division to see if there's anything left-over. Or, which amounts to the same thing, rewrite your numerator with tricks like:

    -x^2.5 = (1-x) x^1.5 - x^1.5


    (Actually, you could have done long division right from the beginning, without bothering with the conjugate -- I wonder why this isn't often taught?)


    Hrm. Maybe the person who wrote the problem expected you to split the numerator into parts that do and do not have fractional powers of x. i.e. as:

    x^.5 (1 - x^2) + (x - x^2)

    I'm not sure if I prefer this, or the division method I mentioned.


    p.s. you meant x^(5/2) not x^5/2.
     
    Last edited: Aug 31, 2006
  6. Aug 31, 2006 #5
    I tried getting it into other formats such as x^.5-x^2/1-x^.5 but it didn't work out too well.
     
  7. Aug 31, 2006 #6
    How exactly would that work out (I'm kinda lost here)? I'm just trying to rationalize the function so as to find the limit for it.
     
  8. Aug 31, 2006 #7

    Hurkyl

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    (I've added a bit more to my previous post)


    Well, division here works just like ordinary division

    Code (Text):


          x^1.5 + x
        ------------------------
    1-x | x^.5 + x - x^2 - x^2.5
          x^1.5 - x^2.5
          ----------------------
          x^.5 + x - x^1.5 - x^2
          x - x^2
          ----------------------
          x^.5 - x^1.5
     
    I've done the first two steps for you -- first, I chose x^1.5, because if I multiply the divisor by x^1.5, I can cancel out the biggest term in the dividend after subtracting.

    (I've decided to have high powers of x be the "biggest" term -- you can do the opposite if you prefer)
     
  9. Aug 31, 2006 #8
    How did you get that?

    Not exactly sure how you got that as well...

    So, what about the remainder resulting from the long division? You're saying that it'll factor into ((1-x)(x^1.5+x))/(1-x). The 1-x will cancel out leaving x^1.5 +x?
     
    Last edited: Aug 31, 2006
  10. Aug 31, 2006 #9
    Yeah, my mistake. Not used to typing it all out on a keyboard.:tongue2:

    Edited.
     
  11. Aug 31, 2006 #10

    Hurkyl

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    Try multiplying it out! This is a very, very useful trick -- you should learn it!

    Again, try multiplying it out!


    (Or did you mean "how did I think of it"?)
     
  12. Aug 31, 2006 #11
    Wait, I see how you got those two parts though I'm not sure how that would help me out though.:uhh:
     
  13. Aug 31, 2006 #12
    Starting from the original equation, it can be solved with partial fractions to be:
    x^1.5-1+(1-x^1.5)/(1-x^0.5)
    =x^1.5-1+(1-x^0.5)*(1+x^0.5+x)/(1-x^0.5)

    Limit as x -> 1 is therefore 3.
     
  14. Aug 31, 2006 #13

    Hurkyl

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    If you stop there, then yes. (Except you swapped the quotient and the remainder):

    (x^.5 + x - x^2 - x^2.5) / (1-x)
    = (x^1.5 + x) + (x^.5 - x^1.5) / (1-x)

    but the division isn't finished -- you can get a better remainder!
     
  15. Aug 31, 2006 #14
    Yeah, I see now but I don't see how that will help me factor the top though...
     
  16. Aug 31, 2006 #15

    Hurkyl

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    I'm suggesting that, if you wanted to do it this way, to try this trick a few times, and see if you wind up with something useful! (This is what I first did)


    Stare at it for a while. Maybe putting it back into conext will help: you want the limit of

    [tex]
    \frac{\sqrt{x} (1 - x^2) + (x - x^2)}{1 - x}
    [/tex]
     
  17. Aug 31, 2006 #16
    Schrodinger's Cat, you should start with the original equation and do the long division. You'll get the answer that I obtained in my earlier reply.
     
  18. Aug 31, 2006 #17
    How did you get that part?
     
  19. Aug 31, 2006 #18
    I did long division of the original equation.
     
  20. Aug 31, 2006 #19
    Ok, but I still don't see how to get rid of the 1-x part.

    I'll finish the long division though I can't remember it all that well.
     
  21. Aug 31, 2006 #20

    Hurkyl

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    What if you were working with:

    [tex]
    \frac{1 - x^2}{1 - x}
    [/tex]
    ?
     
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