How to find the limit of ln⁡(ln⁡(nn−1))ln⁡(n)?How do you findlimn→∞ln⁡(ln⁡(nn−1))ln⁡(n)I know it's -1, but I had to plot it.

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How to find the limit of ln⁡(ln⁡(nn−1))ln⁡(n)?How do you findlimn→∞ln⁡(ln⁡(nn−1))ln⁡(n)I know it&#039;s -1, but I had to plot it.

Cristian Rosales · Accepted Answer

Hint: You can use L&#039;hopital&#039;s rule twice.Added: If you use L&#039;hopital&#039;s rule twice then you get the expression−nn−1.Another approach:ln⁡(ln⁡(nn−1))ln⁡(n)=ln⁡(−ln⁡(1−1n))ln⁡(n)=ln⁡(−(−1n−O(1n2)))ln⁡(n)∼ln⁡(1n)ln⁡(n)=−ln⁡(n)ln⁡(n)=−1.Note that, in the above derivations, we used the Taylor series ofln⁡(1−x)=−x−x22−⋯and the following property of ln⁡(x)ln⁡(1a)=−ln⁡(a).

Adrien Ho · Accepted Answer

Hint:What is limn→∞nn−1Convert the expression into ±∞∞ form. Then you have L&#039;hoptial&#039;s rule.

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