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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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Ahaan Rungta
4 years, 4 months ago
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Top NewestThis 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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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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You made my day , Thx !
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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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'nat' is natural log(base e), Budha.
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0
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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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1
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it is 1 , m sure :)
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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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Its 0
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\(e^{\frac{-1}{2}}\)
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sin33 in surd form
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