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Calculate limit

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data

    [tex]\lim_{x \rightarrow 0} \frac{(1+tan(3x))^{1/3} x^2}{log_2(arcsin(x^3)+1)}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    i tried to use l'hopital twice but i get a very weird function, so there another way to calculate it?
     
  2. jcsd
  3. Feb 5, 2012 #2

    Dick

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    I'd be interested to see these weird functions. l'Hopital is the best way to do it. There is a trick that might help you. lim x->0 (1+tan(3x))^(1/3)=1. That's pretty harmless. I wouldn't include it in the l'Hopital treatment. In general if a factor of your function has a nonzero finite limit, just set it aside and concentrate on the rest.
     
  4. Feb 6, 2012 #3
    thank you Dick,
    i get:
    [tex]Dx^2= 2x[/tex],
    [tex]D log_{2}(arcsin(x^3)+1)=\frac{3x^2}{log2(arcsin(x^3)+1)\sqrt{1-x^6}}[/tex],
    [tex]\lim_{x \to 0} \frac{2x}{\frac{3x^2}{log{2} (arcsin(x^3)+1) \sqrt{1-x^6} } } =
    \frac{2}{3}\lim_{x \to 0} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x} →
    \frac{2}{3}\lim_{x \to 0^-} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x}= \frac{log2}{0^-} = -\infty \vee
    \frac{2}{3}\lim_{x \to 0^+} \frac{log{2} (arcsin(x^3)+1) \sqrt{1-x^6}} {x} = \frac{log2}{0^+}=+\infty
    [/tex]
     
  5. Feb 6, 2012 #4

    Dick

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    Looks ok to me!
     
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