Is x=1 the Solution to nx ~ n ln(n) as n Approaches Infinity?

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SUMMARY

The discussion centers on the mathematical relationship nx ~ n ln(n) as n approaches infinity. The initial assumption that x=1 is incorrect, as it contradicts the behavior of ln(n) in relation to n. The conclusion drawn is that there is no solution to the equation, as the limit of ln(n)/n^k approaches 0 for k > 0 and infinity for k ≤ 0, confirming the absence of a valid x that satisfies the equation.

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How do I solve for x in the relationship below:

nx ~ n ln(n), as n -> infinity

The answer that I'm getting is x=1, but that must be wrong since 1 ~ ln(n) as n-> infinity is wrong.
 
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japplepie said:
How do I solve for x in the relationship below:

nx ~ n ln(n), as n -> infinity

The answer that I'm getting is x=1, but that must be wrong since 1 ~ ln(n) as n-> infinity is wrong.

There is no solution: [tex]\frac{\ln n}{n^k} \to \begin{cases} 0 & k > 0 \\ \infty & k \leq 0 \end{cases}[/tex]
 
oh boy
 

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