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*

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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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 \( k \). – Ahaan Rungta · 3 years, 11 months ago

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– Haroun Meghaichi · 3 years, 11 months ago

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.Log in to reply

– Ahmed Taha · 3 years, 11 months ago

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

– Ricky Escobar · 3 years, 11 months ago

It is a website: www.artofproblemsolving.com.Log in to reply

– Ahmed Taha · 3 years, 11 months ago

You made my day , Thx !Log in to reply

– Piyal De · 3 years, 11 months ago

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

– Budha Chaitanya · 3 years, 11 months ago

what is nat in ln(nat log)?Log in to reply

– Piyal De · 3 years, 11 months ago

'nat' is natural log(base e), Budha.Log in to reply

0 – Vamsi Krishna Appili · 3 years, 11 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, 11 months ago

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1 – Tim Ye · 3 years, 11 months ago

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it is 1 , m sure :) – Aman Rajput · 3 years, 11 months ago

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Can you tell me the ans Ahaan? – Piyal De · 3 years, 11 months ago

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– Ricky Escobar · 3 years, 11 months ago

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

– Piyal De · 3 years, 11 months ago

Yeah, me too.Log in to reply

Its 0 – Shiv Anshu · 3 years, 11 months ago

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\(e^{\frac{-1}{2}}\) – Swapnil Sharma · 3 years, 11 months ago

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sin33 in surd form – Obarewo John · 3 years, 11 months ago

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