# Homework Help: PUHLEAZE help with this limit

1. Oct 13, 2006

### 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 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. Oct 13, 2006

### 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

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

Last edited: Oct 13, 2006
3. Oct 13, 2006

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

4. Oct 13, 2006

### arildno

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

5. Oct 13, 2006

### mohlam12

yeah right... :-s

6. Oct 13, 2006

### Checkfate

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

7. Oct 13, 2006

### mohlam12

umm equals one ?! so ...

8. Oct 14, 2006

### mohlam12

still no one !?

9. Oct 14, 2006

### mohlam12

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

10. Oct 14, 2006

### Checkfate

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

11. Oct 14, 2006

### 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 doesnt help since I am not supposed to use the derivatives

12. Oct 14, 2006

why not look at the graph?

13. Oct 14, 2006

### 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