Help with Limit Problem: Solving Using Various Methods | Expert Tips

[mohlam12]

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


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


okay i tried ti put x=cos^3 (X) but couldn't 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 ...

 

[mohlam12]

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


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

 

[courtrigrad]

Use L'Hopitals rule. \frac{d}{dx} \arctan x = \frac{1}{1+x^{2}}

 

[arildno]

He isn't allowed to know the derivative of arctan..

 

[mohlam12]

yeah right... :-s

 

[Checkfate]

\lim_{y\rightarrow 0}\frac{sin(y)}{y}

 

[mohlam12]

umm equals one ?! so ...

 

[mohlam12]

still no one !?

 

[mohlam12]

here is it...
\lim_{x \rightarrow 1^{-}} \frac{\sqrt[3]{\arctan x} - \arccos \sqrt[3]{x} - \sqrt[3]{\frac{\pi}{4}}}{x-1}

 

[Checkfate]

woops, yea sorry was trying to get your limit to show up... Something weird musta happened. Sorry.

 

[mohlam12]

I realize that this is the derivative of \sqrt[3]{\arctan x} - \arccos \sqrt[3]{x} at the point 1. But I guess that doesn't help since I am not supposed to use the derivatives

 

[courtrigrad]

why not look at the graph?

 

[mohlam12]

okay... here is what i have done so far...
let's put \cos^{3}y = x
so the function becomes:
\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)}
which is equal to
\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}
which is +infinity

okay but there is one thing wrong here... \arccos \sqrt[3]{\cos^{3}y} is not equal to y because \sqrt[3]{\cos^{3}y} 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