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Find all the different limits in + infinity

  1. Sep 16, 2007 #1

    JPC

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

    f(x) = sqroot(ax^2 + 1 ) - x

    find all the different limits in + infinity that f can have with all the different "a" values

    2. Relevant equations



    3. The attempt at a solution

    i dont know if i did something wrong here :

    - if a = 0 then f = 1 - x, limit : - infinty

    - if a < 0 then no f value in + infinite
    , (except if a = ' - 1 / +infnte', limit : - infinte)

    - if a > 0 then

    lim infinite ( f(x) ) = lim +infinite ( root (ax^2 . (1 + 1 / ax^2) ) - x )
    = lim +infinte ( root(a) . x - x)
    = lim +infinte ( (root(a) - 1) x)

    so if a = 1, limit is 0
    if a > 1 , limit is + infinity
    if 0 < a < 1, limit is - infinity


    PS ; if badly wrote, ect, im sorry , but i was on a pc with a wierd keyboard that really thrustrated me
     
  2. jcsd
  3. Sep 16, 2007 #2

    EnumaElish

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    Can you explain "if a < 0 then no f value in + infinite"?

    How do you go from "lim +infinite ( root (ax^2 . (1 + 1 / ax^2) ) - x )" to "lim +infinte ( root(a) . x - x)"?
     
  4. Sep 18, 2007 #3

    JPC

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    hum

    - considering the equation g(x) = ax² + 1
    if a < 0, the, when x reaches +infinite, g(x) reaches -infinite
    and since we only want to study f(x) in R , and not in complex numbers, we dont consider to be a value of f(x) when x reaches +infinite

    (im not sure if you say 'reach' in english , but in french we call it 'tend vers')

    - now if a > 0 :

    lim + infinite ( f(x) )
    = lim +infinite ( sqroot (ax^2 . (1 + 1 / ax^2) ) - x )

    and

    lim + infinite (1 + 1 / ax^2) = 1 + 0 = 1

    so

    lim +infinite ( sqroot (ax^2 . (1 + 1 / ax^2) ) - x )
    = lim +infinite ( sqroot (ax^2 ) - x )
    = lim +infinite ( sqroot(a)* x - x )
    = lim +infinite ( (sqroot(a) -1 ) * x )

    so if
    (sqroot(a) -1 ) > 0
    then
    lim + infinite ( f(x) ) = +infinite

    and if
    (sqroot(a) -1 ) < 0
    then
    lim + infinite ( f(x) ) = -infinite

    finding (sqroot(a) -1 ) = 0
    sqroot(a) = 1
    a = 1

    so if
    a>1
    then
    (sqroot(a) -1 ) > 0
    lim + infinite ( f(x) ) = +infinite

    and if
    0<a<1
    then
    (sqroot(a) -1 ) < 0
    lim + infinite ( f(x) ) = -infinite
     
  5. Sep 19, 2007 #4

    JPC

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    and also :

    its not in my program yet (will be in a few months) but is this true for a < 0 : ???

    since
    lim +oo(f(x)) = lim +infinite ( (sqroot(a) -1 ) * x )

    a = -b , a<0

    sqroot(-b) - 1 = i * sqroot(b) - 1

    so lim +infinite ( (i* sqroot(b) -1 ) * x )

    so lim +oo(f(x)) = (h, +oo)

    - Now lets find that angle h :

    r² = (-1)² + (root(b))² = b+1
    r = root(b+1)
    h = z + pi/2

    by looking at the rectangular triangle of sides : 1, root(b), root(b+1) :

    sin(z) = 1 / root(b+1)

    z = sin^-1(1 / root(b+1))

    so h = pi/2 + sin^-1(1 / root(b+1))

    So when x-> +oo, and a<0 :
    lim lim+oo (f(x) ) = (pi/2 + sin^-1(1 / root(b+1)) , +oo)


    ok, i havent worked at all with complex numbers atm, i just found that with the basic knowledge and understanding i have of complex numbers ( i = root(-1), i² = - 1, and the graphical represantation of complex numbers)

    So , if i did anything wrong here, could you tell me please
     
  6. Sep 19, 2007 #5

    EnumaElish

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    Approaches, tends toward (formal) or goes to (informal).
     
  7. Sep 22, 2007 #6

    JPC

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    yes, but did i do any mistake in what i did ?

    i mean did i use a wrong method , and just got lucky to get the right answer ? (i checked the limits with my calculator)
    Because , what counts more in french schools is the way you find the answer rather than the answer itself
     
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