Compute \[ \lim\limits_{n\to\infty}\left(\sqrt[n]{\log\left|1+\left(\dfrac{1}{n\cdot\log\left(n\right)}\right)^k\right|}\right). \]

*Source: A friend of mine*

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## Comments

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TopNewestThis actually turned out easier than I thought, thanks to Art of Problem Solving user aziiri. This was his solution: \[ \forall\ x\geq 0\ :\ x- \frac{x^2}{2}\leq \ln(1+x)\leq x.\] Apply to get that the limit equals \( \boxed {1} \) for any real number \( k \).

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Yep, that's me. Note that even if we replace \(k\) with \(-n\) we still get \(1\), because the n-th degree root tends towards one a lot faster that the logarithm and the identity function.

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Also you can take ln(nat log) and then evaluate it, it would come ln(L)=0 where 'L' is the limit.

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what is nat in ln(nat log)?

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Can u provide this "Art of Problem Solving" for me ?! is it a book or what ?!

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It is a website: www.artofproblemsolving.com.

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Can you tell me the ans Ahaan?

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Numerically, I have found the answer to be \(1\) for all values of \(k\).

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Yeah, me too.

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it is 1 , m sure :)

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1

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I've never been an expert on limits but i think its 0... if n is realy large, 1/(nlogn) will be 0 pretty much, regardless of the k... then it becomes the nth root of the square root of log(1) which is the nth root of 0 which is 0.. sure theres a way better way of doing this lol

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0

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Its 0

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\(e^{\frac{-1}{2}}\)

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sin33 in surd form

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