Finding a Limit Using L'Hopital's Rule

[tesla93]

Homework Statement

lim x→∞ of ((x^3+2x-∏)^(1/3))-x

Homework Equations

L'Hopital's Rule: If it is an indeterminate form, take the derivative of the top and bottom until you do not get an indeterminate form anymore.

The Attempt at a Solution

lim x→∞ of ((x^3+2x-∏)^(1/3))-x

lim x→∞ (((x^3+2x-∏)^(1/3))-1)/(x^3)/(1/x)

lim x→∞ (1/3(1+2/x^2-∏/x^3)^(2/3))/(-1/x^2)

I do not know how to continue from here. I think I'm missing something in my 3rd step (multiplying by the derivative of the inside function?) Any help would be awesome!

 

[Mark44]

Homework Statement

lim x→∞ of ((x^3+2x-∏)^(1/3))-x

Homework Equations

L'Hopital's Rule: If it is an indeterminate form, take the derivative of the top and bottom until you do not get an indeterminate form anymore.

This is not accurate. L'Hopital's Rule doesn't apply to just any old indeterminate form - only the forms [0/0] or [±∞/∞].

What you have above is the indeterminate form [∞ - ∞], so L'H doesn't apply.

What I would do is to multiply by 1 in the form of something over itself. The "something" would be whatever is needed to come up with the difference of cubes.

The Attempt at a Solution

lim x→∞ of ((x^3+2x-∏)^(1/3))-x

lim x→∞ (((x^3+2x-∏)^(1/3))-1)/(x^3)/(1/x)

lim x→∞ (1/3(1+2/x^2-∏/x^3)^(2/3))/(-1/x^2)

I do not know how to continue from here. I think I'm missing something in my 3rd step (multiplying by the derivative of the inside function?) Any help would be awesome!

 

[tesla93]

Oh okay I understand that. Thanks! I think I've got the right answer now