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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 $k$.

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

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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.

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## Comments

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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 $k$.

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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.

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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.

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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.

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it is 1 , m sure :)

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1

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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

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0

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Its 0

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$e^{\frac{-1}{2}}$

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sin33 in surd form

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