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What could be the limit of this expression? $$\lim_{N \to \infty} \dfrac{\log(N)^3}{\pi N}$$

when we have only $\log(N)$ it gives zero by using a bound, but what can we say about $\log(N)^3$ ?

StubbornAtom
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Lina
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1 Answers1

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tl;dr: It is the same.

You can rewrite (ignoring the $\pi$, which is not changing the result) $$ \frac{(\log N)^3}{N} = \left( \frac{\log N}{N^{1/3}}\right)^3 = \left( \frac{3\log (N^{1/3})}{N^{1/3}}\right)^3 = 27\left( \frac{\log (N^{1/3})}{N^{1/3}}\right)^3 $$ and since $N^{1/3}\xrightarrow[N\to\infty]{}\infty$, we have that $$ \frac{\log (N^{1/3})}{N^{1/3}}\xrightarrow[N\to\infty]{}0 $$ using the limit you already know. By continuity of $x\mapsto x^3$, we then have $$ \frac{(\log N)^3}{N} \xrightarrow[N\to\infty]{}0^3 = 0. $$

Note: This holds for any (constant) exponent: for every fixed $k\in\mathbb{R}$, $$ \frac{(\log N)^k}{N} \xrightarrow[N\to\infty]{}0. $$

Clement C.
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  • Thank you very much. that's what I was looking for. – Lina Feb 28 '17 at 16:04
  • You're welcome. If that settled your question, consider accepting the answer by clicking on the $\checkmark$ on the left. – Clement C. Feb 28 '17 at 16:05
  • Clean and concise (+1). – Mark Viola Feb 28 '17 at 16:16
  • Just curious ... what is the purpose for the "too long;didn't read" preface? What is too long and who didn't read what? Sorry for my confusion here. – Mark Viola Feb 28 '17 at 16:20
  • If the OP wants to know the answer immediately, before getting the proof; or if a reader wants to know why it's alright to discard the factor $\pi$ (which plays no role whatsoever since the limit is $0$ anyway). I'm basically using it to summarize the gist of the answer. @Dr.MV – Clement C. Feb 28 '17 at 16:21