Does limit ln(n)/n^c -> 0 for any c>0?

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In summary, the conversation discusses the question of whether the limit ln(n)/n^c approaches 0 for any c>0. The conversation also mentions different methods for proving this statement, such as using epsilon-delta proofs or BigO SmallO notation. The conversation concludes by mentioning the usefulness of the Bolzano Weierstrasse Theorem in approaching this problem in real analysis.
  • #1
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
 
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  • #2
It is obviously true. There are a lot of ways to prove this. What kind of class is this for? Are you expected to be rigorous in your solutions? N-epsilon proofs and asymptotic behaviour are very different arguments in terms of the development required in an analysis setting.
 
  • #3
Yes, this is for real analysis class
 
  • #4
What tools do you have available? How was ln(n) derived?

There is a nifty theorem that says that if a sequence converges then any subsequence will converge to the same limit. This will enable you to look at ln(x)/x^c in R rather than in N.
 
  • #5
Yes, we can use Bolzano Weierstrasse Theorem. Please feel free to proceed.
My level is on baby Rudin, the this is the first course in real analysis.
 

Related to Does limit ln(n)/n^c -> 0 for any c>0?

1. What is the significance of the limit ln(n)/n^c?

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.

2. Is the limit ln(n)/n^c always equal to 0?

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.

3. Can the limit ln(n)/n^c be evaluated for negative values of n?

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.

4. How is the limit ln(n)/n^c affected by changes in the value of c?

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.

5. Is there a relationship between the limit ln(n)/n^c and the natural logarithm ln(n)?

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.

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