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Limit of 0/0 indeterminant form, n and k for existence of limit |
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| Jun23-12, 12:30 PM | #18 |
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Limit of 0/0 indeterminant form, n and k for existence of limit |
| Jun23-12, 12:35 PM | #19 |
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What I actually need is to find the value of n so I can find k as well using the relation derived earlier. For now, I answered the question through glamorous "hit-and-trial" considering it works best if this problem is to be solved under two minutes. Cancelling 4 with n2 with n = 2 and getting k = 5. (And the limit do exist, 0) |
| Jun23-12, 12:59 PM | #20 |
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Mentor
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There's a BIG difference. |
| Jun23-12, 03:57 PM | #21 |
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Ok, I see your confusion. Let me try.
0/0 in essence is comparing relative order of infinitesimals. Because both numerator and denominator are infinitesimal as x ->0. The relative order of infinitesimal can be analyzed by doing Taylor expansion at point x=0, and look at the *FIRST* surviving term. Fornatutely, in this problem, the first surviving terms pop up pretty quickly. If I have done my math right, the Taylor expansion at x=0 would be: [tex]\dfrac{(\dfrac{5n^2}{4}-k)x^2+\dfrac{5n^4}{64} x^4 +o(x^4)}{-\dfrac{1}{2}x^3+o(x^3)}[/tex] In order for it to have a finite limit, the numerator must not be lower order of infinitesimal than the denominator. That means any lower than cubic power terms in the numerator must die so that numerator is higher order of infinitesimal than the denominator. This is how you get the magic n,k relation. Once the order of both is determined, the actual limit is determined by the ratio of the second (now the first) surviving term in the numerator and the first in the denominator which is not and will never be dependent of n,k as you can see for the numerator. Hope that helps. |
| Jun24-12, 01:18 AM | #22 |
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| Jun24-12, 05:33 AM | #23 |
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