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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
4 years, 4 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 - 4 years, 4 months ago

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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.

Haroun Meghaichi - 4 years, 4 months ago

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

Ahmed Taha - 4 years, 4 months ago

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It is a website: www.artofproblemsolving.com.

Ricky Escobar - 4 years, 4 months ago

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@Ricky Escobar You made my day , Thx !

Ahmed Taha - 4 years, 4 months ago

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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.

Piyal De - 4 years, 4 months ago

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what is nat in ln(nat log)?

Budha Chaitanya - 4 years, 4 months ago

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@Budha Chaitanya 'nat' is natural log(base e), Budha.

Piyal De - 4 years, 4 months ago

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0

Vamsi Krishna Appili - 4 years, 4 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 - 4 years, 4 months ago

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1

Tim Ye - 4 years, 4 months ago

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it is 1 , m sure :)

Aman Rajput - 4 years, 4 months ago

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Can you tell me the ans Ahaan?

Piyal De - 4 years, 4 months ago

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

Ricky Escobar - 4 years, 4 months ago

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Yeah, me too.

Piyal De - 4 years, 4 months ago

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Its 0

Shiv Anshu - 4 years, 4 months ago

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

Swapnil Sharma - 4 years, 4 months ago

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sin33 in surd form

Obarewo John - 4 years, 4 months ago

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