- #1
Macleef
- 30
- 0
What is the solution for a limit problem with an indeterminate form of 0/0?
- Thread starter Macleef
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Homework Help Overview
The discussion revolves around a limit problem that results in an indeterminate form of 0/0. Participants are exploring methods to resolve this limit, particularly focusing on the implications of the indeterminate form and potential techniques to simplify the expression.
Discussion Character
- Exploratory, Assumption checking, Mathematical reasoning
Approaches and Questions Raised
- Participants discuss the initial substitution leading to the 0/0 form and question the effectiveness of various algebraic manipulations. Some suggest using L'Hopital's rule as a potential solution, while others mention specific tricks involving multiplication to simplify the limit.
Discussion Status
The discussion is active, with participants sharing different approaches and questioning the validity of certain methods. There is no explicit consensus on the best approach yet, but several lines of reasoning are being explored, including the application of L'Hopital's rule and algebraic manipulation.
Contextual Notes
Participants are operating under the constraints of the problem's indeterminate form and are considering various mathematical techniques without reaching a definitive solution.
- #2
Mathdope
- 138
- 0
You should have 0/0 not 6/0. In any case, try a trick similar to what you did with the radical in the denominator.
- #3
workerant
- 40
- 0
Well if you plug the four in initially you get:
2-2/(3-3)=0/0
Your trick did not work nor does using the other radical as Mathdope suggests.
But, never fear, 0/0 limits call for one man...
L'Hopital!
Since it is indeterminate form, it is eligible for L'Hopital's rule. That should make it work.
2-2/(3-3)=0/0
Your trick did not work nor does using the other radical as Mathdope suggests.
But, never fear, 0/0 limits call for one man...
L'Hopital!
Since it is indeterminate form, it is eligible for L'Hopital's rule. That should make it work.
- #4
Dick
Science Advisor
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workerant said:
Multiplying by 2+sqrt(x) does so work. You can then cancel the factor that's going to zero in the numerator and the denominator.