# Finding a Limit Using L'Hopital's Rule

1. Apr 11, 2012

### tesla93

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

lim x→∞ of ((x^3+2x-∏)^(1/3))-x

2. Relevant 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.

3. 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!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 11, 2012

### Staff: Mentor

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.

3. Apr 11, 2012

### tesla93

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