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I need help with a couple calculus questions.

Discussion in 'The Lounge' started by colbystephens, Feb 27, 2006.

  1. colbystephens

    colbystephens 1 ton status GMOTM Winner

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    not real tough stuff here. anyway, i'm trying to find the limit of (t[sup]2[/sup]-4)/(t[sup]3[/sup]-8) as t tends to 2. i can't figure out the generic form of a[sup]3[/sup]-b[sup]3[/sup] so i can determine how things cancel out. by generic form i'm referring to, as an example, (a[sup]2[/sup] - b[sup]2[/sup]= (a-b)(a+b).

    also, i'm looking for the limit of (sqrt(x[sup]2[/sup]+x+1)-sqrt(x[sup]2[/sup]-1)) as x tends to positive infinity. when i calculate it, i get + or - 1/2. can someone verify that result for me?

    thanks!
     
  2. colbystephens

    colbystephens 1 ton status GMOTM Winner

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    hmmm. looks like my html code doesn't work. i'll try and fix that. for now, where it says [sup]2[/sup] i'm referring to "squared" or "cubed" depending on the number between the code. thanks!
     
  3. colbystephens

    colbystephens 1 ton status GMOTM Winner

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    ok. i don't know how to format it. sorry.

     
  4. kyser_soze

    kyser_soze 1/2 ton status

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    use this x^2, or x^3
     
  5. colbystephens

    colbystephens 1 ton status GMOTM Winner

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    ok - got the first one figured out. can someone still verify my other question tho?
     
  6. jekquistk5

    jekquistk5 Weld nekid Premium Member GMOTM Winner

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    I'm still confused on your html code or else I could help

    ok give me a couple minutes but from looking at if a little harder I got


    (sqrt(x[sup]2[/sup]+x+1)-sqrt(x[sup]2[/sup]-1)) =

    Lim x->+ infinity sqrt((x^2)+x+1)-sqrt((x^2)-1) and you can use the limit laws to break that up to lim x->+ inifinity sqrt((x^2)+x+1) - Lim x-> sqrt((x^2)-1)

    ahh my brain is fried right now and its been 2 yrs since calc

    Ok you probably dont want to do that because that puts it into an indeterminate form. Good luck sorry I'm no help.. but atleast I tried
     
  7. colbystephens

    colbystephens 1 ton status GMOTM Winner

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    thanks for the try. i found out that my answer is not + or -, it's just + because i'm going for positive infinity.
     
  8. Jagged

    Jagged 1 ton status

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    Second answer... approaches (+)infinity.

    x^2+x+1 > x-1
     
  9. colbystephens

    colbystephens 1 ton status GMOTM Winner

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    i'm pretty sure that's not correct. as x tends to infinity, it should tend to +1/2. but, if you care to explain, i'd be open to objections.
     
  10. Jagged

    Jagged 1 ton status

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    Power rule

    x^n > x^m => lim[x->inf] ( x^n - x^m ) = inf (sqare root of x^2 is x, and x is x^1/2)

    Also:
    lim[x->inf] x^n = ( lim[x->inf] x ) ^ n is true IIRC


    edit:
    Im a dumb****... didnt see the second part is x^2... thought it was x

    oops... might be 1/2 then
     
  11. Jagged

    Jagged 1 ton status

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    Hrm... what's your proof for 1/2?

    I think I can prove infinity with the squeeze theorem.

    I just broke out my calculus book :(
     
  12. colbystephens

    colbystephens 1 ton status GMOTM Winner

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    after multiplying by the conjugate, and cancelling, you end up with x/(sqrt(x^2 + x + 1) + sqrt(x^2 - 1)), then pull out an x^2 from the denominator so that you have x/(sqrt[x^2(1 + (x/x^2) + (1/x^2)] + sqrt[x^2(1 - (1/x^2))]. Then, one can take the root of x^2 so you end up with x/[(x)(sqrt(1 + (x/x^2)+ (1/x^2))+sqrt(1 - (1/x^2)]. Then you separate it so that x/x=1 and you are left with 1/[(sqrt(1 + (x/x^2)+ (1/x^2))+sqrt(1 - (1/x^2)]. Then, factoring infinity in for x, everything with an x denominator effectively goes to 0 and you're left with 1/(sqrt(1)+sqrt(1)) which equals 1/2. Notice that the x in red above should have been pulled out as an absolute value, but since we are only dealing with positive infinity, only the positve x need be considered.
     
  13. az-k5

    az-k5 1/2 ton status

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    I got infinity doing it in my head. What calc is this? I II or III (so I can kinda know what rules you are allowed to use)

    Even if they approach infinity at differing rates, it is going from the positive side so that negates a bunch. The intermidiate found using the limit laws allows for an 'infinity - infinity' situation wich you can L'Hptel (sp = bad) down to infinity - infinty again giving the limit is zero. Do it again and you have infinty. (I am still to lazy to find a pencil and paper so this is just from memory and brain work)
     
  14. colbystephens

    colbystephens 1 ton status GMOTM Winner

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    effectively calc 1
     
  15. Jagged

    Jagged 1 ton status

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    L'Hopital!!!!! I forgot that rule.

    You can prove it with L'Hopital easily

    (the square root really has no affect on the outcome of this problem... just a small nusance for using the proper limit laws)

    Get rid of the x^2
    x+1 > -1
    Infinity
     
  16. colbystephens

    colbystephens 1 ton status GMOTM Winner

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    not familiar with any of that new age hippity-hoppity. care to explain?
     
  17. Jagged

    Jagged 1 ton status

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    Erm.. forgot the whole rule

    2x+1 > 1
    Infinity

    Anywho, I'm gonna take a picture for you
     
  18. thezentree

    thezentree 3/4 ton status

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    That actually made smoke come out of my ears. Seriously, I'm worried it's gonna set the smoke detector off. :eek1:
     
  19. Jagged

    Jagged 1 ton status

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    Note: You can use L'Hostpal recursively....

    lhospital.jpg
     
  20. az-k5

    az-k5 1/2 ton status

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    Most instructors make you earn L'hopital the hard way. It does make limits super easy. Should be in your book, it is probably 2 sections ahead of where you are now.
     

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