Finding a Limit Using L'Hopital's Rule

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SUMMARY

The discussion focuses on evaluating the limit as x approaches infinity for the expression ((x^3 + 2x - π)^(1/3)) - x using L'Hopital's Rule. Participants clarify that L'Hopital's Rule is applicable only for the indeterminate forms [0/0] or [±∞/∞], while the given limit is of the form [∞ - ∞]. The recommended approach involves multiplying by a form of one to facilitate the difference of cubes, allowing for further simplification. Ultimately, the correct application of algebraic manipulation leads to the resolution of the limit.

PREREQUISITES
  • Understanding of limits and infinity in calculus
  • Familiarity with L'Hopital's Rule and its conditions
  • Knowledge of algebraic manipulation techniques, specifically the difference of cubes
  • Basic differentiation skills in calculus
NEXT STEPS
  • Study the conditions for applying L'Hopital's Rule in detail
  • Learn about the difference of cubes and its applications in limits
  • Practice solving limits involving indeterminate forms
  • Explore advanced limit techniques, including algebraic manipulation and series expansion
USEFUL FOR

Students studying calculus, particularly those tackling limits and indeterminate forms, as well as educators seeking to clarify the application of L'Hopital's Rule.

tesla93
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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!
 
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tesla93 said:

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.
tesla93 said:

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

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