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Few calc limit questions need help!

  1. Oct 11, 2012 #1
    few calc limit questions!! need help!

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

    limit x-> -1 of cubed rt (3x-5/25x-2)
    not sure how to go about this at all

    i think the answer is 2/3 but i cant work it up


    lim x->0 √(3x+2) - √2
    x


    r(x)= |3x| A) lim x->0 of r(x) question 2 B) r(0)
    x





    3. The attempt at a solution
     
  2. jcsd
  3. Oct 11, 2012 #2
    Re: few calc limit questions!! need help!

    these werent formatted right. the middle one the x is underneath the top term as well as for the absolute (|3x|)/x

    also, i tried to multiply the middle one by its conjugate √(3x+2) - √2

    i wind up getting (3x)/(x (√(3x+2) - √2) ) i think thats the right track but im not sure
     
  4. Oct 11, 2012 #3

    tiny-tim

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    welcome to pf!

    hi venom2121! welcome to pf! :smile:

    (btw, never reply to your own first post :redface:, use the EDIT button instead … then you'll stay on the No-replies list! :wink:)
    but it isn't 0/0, so what's the difficulty? :confused:

    just put x = -1 ! :rolleyes:
    use a binomial expansion (and √(3x+2) = √x√(3 + 2/x)) :wink:
    try drawing a graph :wink:
     
  5. Oct 11, 2012 #4
    Re: few calc limit questions!! need help!

    the cubed rt problem the answer is supposed to be 2/3 DERP ok had a brain fart haha got that now..

    im still confused about the binomial expansion.


    and as far as the absolute value one im not sure. i see that the y values max out at -3 and 3 and as it goes to zero the y values stay at 3 until zero. so is that the limit? would that mean that r(0)=0? also how would you figure this out without graphing?
     

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    Last edited: Oct 11, 2012
  6. Oct 11, 2012 #5

    tiny-tim

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    can't you do the binomial expansion of (1 + x)1/2 ?
    not following you :confused:
    i wouldn't!
     
  7. Oct 11, 2012 #6

    Ray Vickson

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    Re: few calc limit questions!! need help!

    Your "cube-root" problem, as written, is
    [tex] \lim_{x \rightarrow -1} \sqrt[3]{\left( 3x - \frac{5}{25x} - 2\right)} = -\frac{\sqrt[3]{600}}{5},[/tex]
    but perhaps you meant
    [tex] \lim_{x \rightarrow -1} \sqrt[3]{\left( \frac{3x-5}{25x - 2}\right)}.[/tex]
    In that case you should have used brackets, and written cube rt ((3x-5)/(25x-2)).

    RGV
     
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