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
Waste less time on Facebook — follow Brilliant.
No vote yet
18 votes
Comments
Sort by:
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 \( k \). – Ahaan Rungta · 4 years, 1 month ago
Log in to reply
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, 1 month ago
Log in to reply
Can u provide this "Art of Problem Solving" for me ?! is it a book or what ?! – Ahmed Taha · 4 years, 1 month ago
Log in to reply
It is a website: www.artofproblemsolving.com. – Ricky Escobar · 4 years, 1 month ago
Log in to reply
You made my day , Thx ! – Ahmed Taha · 4 years, 1 month ago
Log in to reply
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, 1 month ago
Log in to reply
what is nat in ln(nat log)? – Budha Chaitanya · 4 years, 1 month ago
Log in to reply
'nat' is natural log(base e), Budha. – Piyal De · 4 years, 1 month ago
Log in to reply
0 – Vamsi Krishna Appili · 4 years, 1 month ago
Log in to reply
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, 1 month ago
Log in to reply
1 – Tim Ye · 4 years, 1 month ago
Log in to reply
it is 1 , m sure :) – Aman Rajput · 4 years, 1 month ago
Log in to reply
Can you tell me the ans Ahaan? – Piyal De · 4 years, 1 month ago
Log in to reply
Numerically, I have found the answer to be \(1\) for all values of \(k\). – Ricky Escobar · 4 years, 1 month ago
Log in to reply
Yeah, me too. – Piyal De · 4 years, 1 month ago
Log in to reply
Its 0 – Shiv Anshu · 4 years, 1 month ago
Log in to reply
\(e^{\frac{-1}{2}}\) – Swapnil Sharma · 4 years, 1 month ago
Log in to reply
sin33 in surd form – Obarewo John · 4 years, 1 month ago
Log in to reply