# Homework Help: Tricky limit

1. Nov 8, 2012

### mtayab1994

1. The problem statement, all variables and given/known data

lim x(pi-2arctan(x^(2/3)))
x->-∞

3. The attempt at a solution

I don't know how to remove undefined form. Can someone give me a hint? Thanks in advance.

2. Nov 8, 2012

### micromass

l'Hopitals theorem?

3. Nov 8, 2012

### mtayab1994

I can't use that so can i do xpi-2xarctan(x^(2/3)) which will become:

xpi-[(2arctan(x^2/3)/x^(2/3)]*x(^5/3) , but even that still gives an undefined form.

4. Nov 8, 2012

### micromass

Why can't you use it?

5. Nov 8, 2012

### mtayab1994

Not aloud to use it. Our teacher says it's just used to check if your limit is correct. Plus we still haven't learned how to derive the arctan function.

Last edited: Nov 8, 2012
6. Nov 8, 2012

### mtayab1994

Is there any other way on how to factor it so you can put t=arctan(x^(2/3)). My attempt still gave me -∞+∞ which is undefined.

7. Nov 8, 2012

### Staff: Mentor

Which properties of the arctan function are available?

This would allow to replace the x with some function of t. How does your new limit look like?

8. Nov 8, 2012

### mtayab1994

Well i did something else here is what i did.

$$\lim_{x\rightarrow-\infty}x\pi-\frac{arctan(\sqrt[3]{x^{2}})}{\sqrt[3]{x^{2}}}\cdot2\sqrt[3]{x^{5}}$$

9. Nov 9, 2012

### Staff: Mentor

Well, those terms all diverge, so it does not look very useful.

10. Nov 9, 2012

### mtayab1994

Yes that is what it looks like to me. Do you know what I can use?

11. Nov 9, 2012

### haruspex

Try mfb's suggestion to get rid of the arctan. Then do another substitution so that the variable tends to 0 from above. That will make things clearer.