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I know following limit is eventually equal to $0$ but can someone actually show me the steps for general value of k: $$\lim\limits_{n \to \infty} \dfrac{(\log n)^k}{n} =0$$

I know that attempting to take the limit of the numerator and the denominator just leads to the $\infty/\infty$ indeterminate form so the L'Hopital's Rule must be used taking derivative of numerator and denominator. I don't know how many times I should take it for the general case.

Edit: Is this accurate for a general form of this limit?

$$[k(k-1)(k-2)...3*2*1]\lim\limits_{n \to \infty}\dfrac{1}{n} = 0$$

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Hint. First use L'Hopital's rule to evaluate $$\lim_{n\to\infty}\frac{\log n}{n^{1/k}}$$ for $k>0$.

(Note that if you want the limit for $k=0$ or for $k<0$ then it is not an indeterminate form and is much easier.)

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  • $\begingroup$ Could you explain how you converted my original problem to this first ? $\endgroup$ – Gdgames Gamers Oct 7 '14 at 0:30
  • $\begingroup$ If you get this limit goes to 0, then raising it ot the k'th power also goes to 0, because a continuous function (raising to the k'th power) can commute accross limits $\endgroup$ – Alan Oct 7 '14 at 0:37
  • $\begingroup$ @quickCoder I took the $k$th root, and I did this because then I have a power of $n$, which is pretty easy, instead of a power of $\log n$ which is harder. After finding this limit, take the $k$th power. $\endgroup$ – David Oct 7 '14 at 0:43
  • $\begingroup$ @David Is this a correct way of writing the general form? : $[k( k - 1) (k - 2)(k-3)...3*2*1 ]$$\lim_{n\to\infty}$ $1/n = 0$ $\endgroup$ – Gdgames Gamers Oct 7 '14 at 2:36
  • $\begingroup$ @quickCoder Yes I guess... but I would have thought that was the hard way of doing it ;-) $\endgroup$ – David Oct 7 '14 at 3:08
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Maybe You can try in this way:

Let $t=\log n$. Then $n=10^t$. So the $\lim_{t\to \infty}\frac{t^k}{10^t} $.

Is this more clear?

You should take derivative of $\frac{t^k}{10^t}$ until $(k-)(k-2)\dots (k-m)\dots <0$

May it helps.

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other way:

$\log (n)^k = \log(n^k)$

then $\lim\limits_{n \to \infty} \dfrac{\log (n^k)}{n} = \lim\limits_{n \to \infty} \dfrac{1}{n} .\log (n^k) =\lim\limits_{n \to \infty} \log((n^k)^ \dfrac{1}{n})= \log(\lim\limits_{n \to \infty}(n^k)^ \dfrac{1}{n})$

then I think we know that $\lim\limits_{n \to \infty}(n^k)^ \dfrac{1}{n} = 1$

then our limit is $\log(1) = 0$

or to avoid use L'Hospital's who know how many time.. do:

$\log(n)^k = k.\log(n)$ then use $ \lim\limits_{n \to \infty}\dfrac{(\log n)^k}{n} = \lim\limits_{n \to \infty} k.\dfrac{(\log n)}{n} = k.\lim\limits_{n \to \infty} \dfrac{(\log n)}{n}$ and then our problem it's $\lim\limits_{n \to \infty} \dfrac{(\log n)}{n}$ and to solve it, use L'Hospital once.

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