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[Challenge] Logarithmic Limit Barrage

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

Note by Ahaan Rungta
3 years, 9 months ago

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This 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 \( k \). Ahaan Rungta · 3 years, 9 months ago

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@Ahaan Rungta 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. Haroun Meghaichi · 3 years, 9 months ago

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@Ahaan Rungta Can u provide this "Art of Problem Solving" for me ?! is it a book or what ?! Ahmed Taha · 3 years, 9 months ago

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@Ahmed Taha It is a website: www.artofproblemsolving.com. Ricky Escobar · 3 years, 9 months ago

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@Ricky Escobar You made my day , Thx ! Ahmed Taha · 3 years, 9 months ago

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@Ahaan Rungta Also you can take ln(nat log) and then evaluate it, it would come ln(L)=0 where 'L' is the limit. Piyal De · 3 years, 9 months ago

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@Piyal De what is nat in ln(nat log)? Budha Chaitanya · 3 years, 9 months ago

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@Budha Chaitanya 'nat' is natural log(base e), Budha. Piyal De · 3 years, 9 months ago

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0 Vamsi Krishna Appili · 3 years, 9 months ago

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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 Jord W · 3 years, 9 months ago

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1 Tim Ye · 3 years, 9 months ago

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it is 1 , m sure :) Aman Rajput · 3 years, 9 months ago

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Can you tell me the ans Ahaan? Piyal De · 3 years, 9 months ago

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@Piyal De Numerically, I have found the answer to be \(1\) for all values of \(k\). Ricky Escobar · 3 years, 9 months ago

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@Ricky Escobar Yeah, me too. Piyal De · 3 years, 9 months ago

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Its 0 Shiv Anshu · 3 years, 9 months ago

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\(e^{\frac{-1}{2}}\) Swapnil Sharma · 3 years, 9 months ago

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sin33 in surd form Obarewo John · 3 years, 9 months ago

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