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Yarrrgh I'm helping this guy with calc 1 and apparently I suck now

  1. Sep 15, 2006 #1
    Limits:( If it were things like derivatives and integrals I'd be good to go but nooooo, stupid limits

    lim(x-->0) [(1+ax)^(1/3)-1]/x=1

    find a such that this is true. Guessing on my calculator tells me its 3 but I dunno how to do it analytically. My first guess is clear the radical, but I don't know how to clear a cube root without making it worse, if you can

    lim(x-->2)[sqrt(6-x)-2]/[sqrt(3-x)-1]=?

    again, calculator tells me 1/2, but I can't. Trying to clear either numerator or denominator of its radicals gets either the top or bottom to be 2-x and the other side to be a bigger mess which I can decipher.

    and finally, pick two functions f(x) and g(x) such that lim(x-->0)[f(x)+g(x)] exists, but f(x) and g(x)'s limits as x-->0 DON'T exist

    I really don't think that's possible and wouldn't know where to start finding those functions if it weren't
     
  2. jcsd
  3. Sep 15, 2006 #2
    Oh, and no derivatives yet, so no l'Hopital's Rule
     
  4. Sep 15, 2006 #3

    J77

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    Expand the cube root bit as a binomial series.
     
  5. Sep 15, 2006 #4
    Actually I just figured that one out

    First step is multiply out that x, then subtract the 1, cube both sides, and everything falls into place from there

    the second one has me MEGAstumped though
     
  6. Sep 15, 2006 #5

    J77

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    I would get rid of the square root term on the top...

    I'll let you have a go at doing this.
     
  7. Sep 15, 2006 #6
    Getting rid of either square root doesn't seem to help. Like I said you get 2-x on the side you do it on and some ridiculous string of stuff that still ends up as 0.
     
  8. Sep 15, 2006 #7

    Galileo

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    Like you saud, if you try to get rid if the square root above, you get 2-x in the numerator. If you try to get rid of the square root below, you also get 2-x, but in the denominator. That should give you an idea on what to do. What if you got 2-x on both sides?
     
  9. Sep 15, 2006 #8
    The third one is trivial. Take some f with no limit as x->0, say sin(1/x), and let g = -f :tongue2:
     
  10. Sep 15, 2006 #9
    I KNEW it

    always happens, problems where my first instinct is "hey that's not possible" usually have really simple solutions.
     
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