[Challenge] Logarithmic Limit Barrage

Compute limn(log1+(1nlog(n))kn). \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
6 years, 4 months ago

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18 votes

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Comments

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This actually turned out easier than I thought, thanks to Art of Problem Solving user aziiri. This was his solution:  x0 : xx22ln(1+x)x. \forall\ x\geq 0\ :\ x- \frac{x^2}{2}\leq \ln(1+x)\leq x. Apply to get that the limit equals 1 \boxed {1} for any real number k k .

Ahaan Rungta - 6 years, 4 months ago

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Yep, that's me. Note that even if we replace kk with n-n we still get 11, because the n-th degree root tends towards one a lot faster that the logarithm and the identity function.

Haroun Meghaichi - 6 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 - 6 years, 4 months ago

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what is nat in ln(nat log)?

Budha Chaitanya - 6 years, 4 months ago

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

Piyal De - 6 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 - 6 years, 4 months ago

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It is a website: www.artofproblemsolving.com.

Ricky Escobar - 6 years, 4 months ago

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@Ricky Escobar You made my day , Thx !

Ahmed Taha - 6 years, 4 months ago

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Can you tell me the ans Ahaan?

Piyal De - 6 years, 4 months ago

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Numerically, I have found the answer to be 11 for all values of kk.

Ricky Escobar - 6 years, 4 months ago

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Yeah, me too.

Piyal De - 6 years, 4 months ago

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it is 1 , m sure :)

Aman Rajput - 6 years, 4 months ago

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1

Tim Ye - 6 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 - 6 years, 4 months ago

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0

Vamsi Krishna Appili - 6 years, 4 months ago

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Its 0

Shiv Anshu - 6 years, 4 months ago

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e12e^{\frac{-1}{2}}

Swapnil Sharma - 6 years, 4 months ago

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sin33 in surd form

Obarewo John - 6 years, 4 months ago

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