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Find the limit as x tends to 0

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data

    find the limit as x tends to 0:
    lim x^2/(5th-root(1+5x)-(1+x))

    2. Relevant equations



    3. The attempt at a solution

    Let y=5th-root(1+5x), so now the limit tends to 1 and the limit is now:

    lim (y^5-1)^2/(25(y-1-y^5/5+1/5)

    = lim (y^10-2y^5+1)/(25y-25-5y^2+5)

    This is where i got stuck.
    any help would be very much appreciated.

    (Note, i dont want to use L'hospital's rule)
     
  2. jcsd
  3. Nov 23, 2008 #2
    Re: limit



    Too bad, it would be really easy with l'hospital's rule, you could plug in y=1+h, converting the limit back to h->0. But expanding the y^10 might be tedious:smile: Then you can cancel some h's and see where it goes.
     
  4. Nov 23, 2008 #3
    Re: limit

    Yes it would be much easier to use L'hospital's rule that's why my teacher prohibitted it.

    why would i put y=1+h if i have to expand the power 10?
    Surely there must be a simpler way??
     
  5. Nov 23, 2008 #4
    Re: limit

    You can also do a power series expansion of (1+5x)1/5 around 0....this would be 1+x+...+o(x3)....The first two terms will cancel, so you will need to keep the third one as well...
     
  6. Nov 23, 2008 #5
    Re: limit

    Okay, i'll try that but i didnt think i would need to use the taylor expansion.
     
  7. Nov 23, 2008 #6

    lurflurf

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    Re: limit

    = lim (y^10-2y^5+1)/(25y-25-5y^2+5)
    should be
    = lim (y^10-2y^5+1)/(25y-25-5y^5+5)
    simplify to
    = lim (y^10-2y^5+1)/(y^5-5y+4)/-5
    now factor (hint (y-1)^2 divides the numerator and denominator)
    however it may be easier to use
    z=(-1+y)^2=[-1+5th-root(1+5x)]^2
     
  8. Nov 23, 2008 #7
    Re: limit

    are you sure that (y-1)^2 is a factor????
     
  9. Nov 23, 2008 #8

    lurflurf

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    Re: limit

    yes

    (y^5-1)=(y-1)(y^4+y^3+y^2+y+1)
    (y^5-5y+4)=(y-1)^2(y^3+2y^2+3y+4)

    use synthetic division or long division to see
     
  10. Nov 23, 2008 #9
    Re: limit

    the second one i agree. but; where did (y^5-1) come from?
    the numerator is: y^10-2y^5+1
    now that doesnt divide (y-1)^2
     
  11. Nov 23, 2008 #10

    lurflurf

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    Re: limit

    y^10-2*y^5+1=(y^5-1)^2
     
  12. Nov 23, 2008 #11
    Re: limit

    oh right, sorry.
    but still, (y^5-1)^2 divides y-1 and the denominator divides (y-1)^2
    when we cancel this out we get:
    lim (y^4+y^3+y^2+y+1)/(-5)(y-1)(y^3+2y^2+3y+4)

    so we still cant substitute y=1 to find the limit since the denominator would be 0.
     
  13. Nov 23, 2008 #12

    lurflurf

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    Re: limit

    since the numerator is square all the factors appear twice
    y^10-2*y^5+1=(y^5-1)^2=(y-1)^2(y^3+2y^2+3y+4)^2

    lim (y^10-2y^5+1)/(y^5-5y+4)/-5
    lim (y^5-1)^2/(y^5-5*+4)/-5
    lim [(y^4+y^3+y^2+y+1)(y-1)]^2/[(-5)(y-1)^2(y^3+2y^2+3y+4)]
    so that your
    lim (y^4+y^3+y^2+y+1)/(-5)(y-1)(y^3+2y^2+3y+4)
    should be
    lim (y^4+y^3+y^2+y+1)^2/[(-5)(y^3+2y^2+3y+4)]
     
  14. Nov 23, 2008 #13
    Re: limit

    Is this really easier than using a power expansion of this fifth root...?
     
  15. Nov 23, 2008 #14
    Re: limit

    Oh right, that makes sense wat a silly mistake i made!!
    Thanks.

    i dont like power expansions of high powers.
     
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