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Limit of the function

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data
    hello, I have a little problem with some limit of the functions, could you help me? please



    2. Relevant equations

    [tex]\lim_{x \to 0} \quad \displaystyle\frac{\sqrt[3]{1+\mbox{arctg} 3x}-\sqrt[3]{1-\arcsin 3x}}{\sqrt{1-\arcsin 2x}-\sqrt{1+\mbox{arctg} 2x}}
    [/tex]

    3. The attempt at a solution
    it's one of the hardest example from my book, i don't even know how to start it.
    i will be very grateful for help

    and.. i don't know why my tex doesnt work. do you have some idea?
     
    Last edited: Nov 23, 2009
  2. jcsd
  3. Nov 23, 2009 #2

    Mark44

    Staff: Mentor

    You were missing the [ tex] and [ /tex] tags (without the leading spaces inside the brackets).
     
  4. Nov 23, 2009 #3

    Mark44

    Staff: Mentor

    Try multiplying by 1 in the form of (1 + arctan(3x))^(2/3) + (1 + arctan(3x))^(1/3)(1 - arcsin(3x))^(1/3) + (1 - arcsin(3x))^(2/3) over itself.

    The underlying idea is that (a - b)(a^2 + ab + b^2) = a^3 - b^3.
     
  5. Nov 23, 2009 #4
    i did just like you said and i got " (0*2)/[0(1+0+0)] :( that means that this is equal 0? right?

    i typed this: ((1+(arc tg3x))^1/3 -(1+(arc sin3x))^1/3)/((1-(arc sin2x))^1/2-(1+(arc tg2x))^1/2) into wolfram alpha and there is too 0, but even that i don't think that my calculation is good in mathematician meaning.

    what can i do else, what i do wrong?
    any idea?
     
  6. Nov 23, 2009 #5

    Mark44

    Staff: Mentor

    [0/0] is an indeterminate form, which means that an expression that has this form can have any limiting value. Give me a while to take a closer look at this.
     
  7. Nov 23, 2009 #6
    [tex]\frac{\sqrt[3]{1+\arctan 3x}-\sqrt[3]{1-\arcsin 3x}}{\sqrt{1-\arcsin 2x}-\sqrt{1+\arctan 2x}}=\\=\frac{\arctan 3x+\arcsin 3x}{-\arcsin 2x-\arctan 2x} \cdot \frac{\sqrt{1-\arcsin 2x}+\sqrt{1+\arctan 2x}}{\sqrt[3]{(1+\arctan 3x)^2}+\sqrt[3]{(1+\arctan 3x)(1-\arcsin 3x)}+\sqrt[3]{(1-\arcsin 3x)^2}}[/tex]

    so we got -3/2 *2/3 and that equals -1? is that correct? i have to be sure
     
    Last edited: Nov 23, 2009
  8. Nov 23, 2009 #7
    -1 is the correct limit.
     
  9. Nov 23, 2009 #8
    yeap, i know that this is the correct limit, but is this a correct calculation :-> -3/2 *2/3->-1
     
  10. Nov 23, 2009 #9

    Mark44

    Staff: Mentor

    -3/2 * 2/3 = -1, yes. Is that what you're asking?
     
  11. Nov 23, 2009 #10

    Mark44

    Staff: Mentor

    Looks good.
     
  12. Nov 24, 2009 #11
    by -3/2 i was meaning the first statement of the above equation... :]

    but.. i was always thinking that arcus sin3x is going to 0, lim x->0 arcus tg is going to 0 too?
    why here we doesn't the same : arcussin3x->0, arcustg3x->0 -->> 0/0 in first statement.
     
  13. Nov 24, 2009 #12

    Mark44

    Staff: Mentor

    Because 0/0 is not a number. It is one of several indeterminate forms, which means that a limit having this form can come out to be anything. For example, as x --> 0, x/(2x) --> 1/2, x^2/x --> infinity, and x/x^3 --> 0. All three of these limits are of the form [0/0] and their limiting values are different.
     
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