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grossgermany
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Homework Statement
Does limit ln(n)/n^c -> 0 for any c>0?
Homework Equations
The Attempt at a Solution
I wonder if there is an
1.Epsilon Delta Proof
2.Proof using BigO SmallO notation.
Thanks
The limit ln(n)/n^c represents the growth rate of the natural logarithm compared to a polynomial function. It is commonly used in the analysis of algorithms and in studying the rate of growth of a sequence.
No, the limit ln(n)/n^c is only equal to 0 for values of c greater than 1. For values of c less than or equal to 1, the limit approaches infinity as n approaches infinity.
No, the natural logarithm ln(n) is only defined for positive values of n. Therefore, the limit ln(n)/n^c cannot be evaluated for negative values of n.
The limit ln(n)/n^c is inversely proportional to the value of c. This means that as c increases, the limit decreases and approaches 0. As c decreases, the limit increases and approaches infinity.
Yes, the limit ln(n)/n^c is equivalent to the natural logarithm ln(n) when c=1. This is because ln(n)/n^1 reduces to ln(n). However, for any other value of c, the limit ln(n)/n^c is a different function with a different rate of growth.