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Homework Help: PUHLEAZE help with this limit

  1. Oct 13, 2006 #1
    okay, here I have a problem with this limit, i used every method i know of and could solve it... any help or something to get started with be appreciated


    [tex] \lim_{\x\rightarrow 1^{-}\} \frac{\sqrt[3]{\arctan(x)} - \arccos(\sqrt[3]{x}) - \sqrt[3]{frac{\pi}{4}}{x-1}[/tex]


    okay i tried ti put x=cos^3 (X) but couldnt get to a result
    I tried to use the x^3 - y^3 = (x-y)(x²+xy+y²) but no result
    i tried everything :-S

    PS: we didnt learn the derivative function of arctanx ...
     
  2. jcsd
  3. Oct 13, 2006 #2
    i think the latex hasnt worked... it was :
    the limit when x tends to 1- of :
    [arctan(x)]^(1/3) - arccos [x^(1/3)] - (pi/4)^(1/3)
    -----------------------------------------------------
    x-1


    [tex] \lim_{\x\rightarrow 1^{-}\} \frac{\sqrt[3]{\arctan(x)}-\arccos(\sqrt[3]{x}) - \sqrt[3]{frac{\pi}{4}}{x-1}[/tex]
     
    Last edited: Oct 13, 2006
  4. Oct 13, 2006 #3
    Use L'Hopitals rule. [tex] \frac{d}{dx} \arctan x = \frac{1}{1+x^{2}} [/tex]
     
  5. Oct 13, 2006 #4

    arildno

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    He isn't allowed to know the derivative of arctan..:frown:
     
  6. Oct 13, 2006 #5
    yeah right... :-s
     
  7. Oct 13, 2006 #6
    [tex]\lim_{y\rightarrow 0}\frac{sin(y)}{y}[/tex]
     
  8. Oct 13, 2006 #7
    umm equals one ?! so ...
     
  9. Oct 14, 2006 #8
    still no one !? :confused:
     
  10. Oct 14, 2006 #9
    here is it...
    [tex]\lim_{x \rightarrow 1^{-}} \frac{\sqrt[3]{\arctan x} - \arccos \sqrt[3]{x} - \sqrt[3]{\frac{\pi}{4}}}{x-1}[/tex]
     
  11. Oct 14, 2006 #10
    woops, yea sorry was trying to get your limit to show up... Something weird musta happened. Sorry.
     
  12. Oct 14, 2006 #11
    I realize that this is the derivative of [tex] \sqrt[3]{\arctan x} - \arccos \sqrt[3]{x} [/tex] at the point 1. But I guess that doesnt help since I am not supposed to use the derivatives
     
  13. Oct 14, 2006 #12
    why not look at the graph?
     
  14. Oct 14, 2006 #13
    okay... here is what i have done so far........
    let's put [tex]\cos^{3}y = x[/tex]
    so the function becomes:
    [tex]\lim_{y \rightarrow 1^{-}} \frac{\sqrt[3]{\arctan \cos^{3}y} - \arccos \sqrt[3]{\cos^{3}y} - \sqrt[3]{\frac{\pi}{4}}}{(\cos^{3}y-1)}[/tex]
    which is equal to
    [tex]\lim_{y \rightarrow 1^{-}} \frac{\arctan \cos^{3}y - \frac{\pi}{4}} {(\cos^{3}y-1)(\sqrt[3]{\arctan \cos^{3}y}^{2} + \sqrt[3]{\frac{\pi}{4}}^{2} + \sqrt[3]{\arctan \cos^{3}y} \sqrt[3]{\frac{\pi}{4}})} - \frac{y}{\cos^{3}y-1}[/tex]
    which is +infinity

    okay but there is one thing wrong here... [tex]\arccos \sqrt[3]{\cos^{3}y}[/tex] is not equal to y because [tex]\sqrt[3]{\cos^{3}y}[/tex] should be in the interval (0,pi), but actually, it's on the interval of (-pi/2 , pi/2) |because when cos^3y is positive only between -pi/2 and pi/2.
    know what I'm sayin :confused:
     
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