Solution: Solving a Tricky Limit Problem

[mtayab1994]

Homework Statement

lim x(pi-2arctan(x^(2/3)))
x->-∞

The Attempt at a Solution

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

 

[micromass]

l'Hopitals theorem?

 

[mtayab1994]

l'Hopitals theorem?

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.

 

[micromass]

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.

Why can't you use it?

 

[mtayab1994]

Why can't you use it?

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.

 

[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.

 

[mfb]

Which properties of the arctan function are available?

t=arctan(x^(2/3))

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

 

[mtayab1994]

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?

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}}

 

[mfb]

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

 

[mtayab1994]

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

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

 

[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.