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Comments
This actually turned out easier than I thought, thanks to Art of Problem Solving user aziiri. This was his solution: ∀x≥0:x−2x2≤ln(1+x)≤x.
Apply to get that the limit equals 1 for any real number k.
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.
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
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
This actually turned out easier than I thought, thanks to Art of Problem Solving user aziiri. This was his solution: ∀ x≥0 : x−2x2≤ln(1+x)≤x. Apply to get that the limit equals 1 for any real number k.
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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.
Also you can take ln(nat log) and then evaluate it, it would come ln(L)=0 where 'L' is the limit.
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what is nat in ln(nat log)?
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Can u provide this "Art of Problem Solving" for me ?! is it a book or what ?!
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It is a website: www.artofproblemsolving.com.
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Can you tell me the ans Ahaan?
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Numerically, I have found the answer to be 1 for all values of k.
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Yeah, me too.
it is 1 , m sure :)
1
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
0
Its 0
e2−1
sin33 in surd form