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How do take this limit:

$$ \lim_{n\to\infty} \frac{\sqrt{n}}{\log(n)}$$

I have a feeling that it is infinity, but I'm not sure how to prove it. Should I use L'Hopitals Rule?

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It would be interesting to try using Laplace transform and final value theorem: $\lim_{t \to \infty} f(t) = \lim_{s \to 0} \mathcal{L}[f(t)](s).$ – user2468 Mar 30 '12 at 5:13
Other folks have answered this adequately, but it might be worth pointing out that $$\lim_{n\to\infty}{n^c\over\log n} = \infty$$ for all positive $c$. – MJD Mar 30 '12 at 7:20
up vote 7 down vote accepted

Let $n = e^x$. Note that as $n \rightarrow \infty$, we also have $x \rightarrow \infty$. Hence, $$\lim_{n \rightarrow \infty} \frac{\sqrt{n}}{\log(n)} = \lim_{x \rightarrow \infty} \frac{\exp(x/2)}{x}$$ Note that $\displaystyle \exp(y) > \frac{y^2}{2}$, $\forall y > 0$ (Why?). Hence, we have that $$\lim_{n \rightarrow \infty} \frac{\sqrt{n}}{\log(n)} = \lim_{x \rightarrow \infty} \frac{\exp(x/2)}{x} \geq \lim_{x \rightarrow \infty} \frac{\frac{x^2}{8}}{x} = \lim_{x \rightarrow \infty} \frac{x}{8} = \infty$$

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Wouldn't it be sufficient to use the series expansion of $e^{x/2} = \displaystyle\sum_{i=0}^{\infty} \frac{{x}^{i}}{2^i i!}$ and say $\displaystyle\lim_{x\to \infty} \frac{e^{x/2}}{x} = $ $\displaystyle\lim_{x\to \infty} \displaystyle\sum_{i=0}^{\infty} \frac{x^{i-1}}{2^{i-1}i!} = \infty$? – user2468 Mar 30 '12 at 15:24
Ops. I've just notices you used $O(x^3)$ expansion of $e^x.$ Ignore my previous comment. – user2468 Mar 30 '12 at 15:27

Let $a(n) = \frac{\sqrt{n}}{\log(n)}$. We want to show that $a(n)$ grows arbitrarily large.

$a(n^2) = \frac{n}{2\log(n)}$ so $\frac{a(n^2)}{a(n)} = \frac{\sqrt{n}}{2}$, so, for $n >16$, $\frac{a(n^2)}{a(n)} > 2$.

Iterating or inducting or multiplying, $\frac{a(n^{2^k})}{a(n)} > 2^k$, so $a(n)$ gets arbitrarily large.

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The hypotheses for l'Hopital are met, so why not try it and see?

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Wouldn't this fit better as a comment? – Pedro Tamaroff Mar 30 '12 at 0:25
The question was, "should I use l'Hopital's Rule?" I could have simply answered, "yes", but 1) that's too short for an answer, and 2) on pedagogic grounds I'd rather see OP work out whether l'H works than give the game away myself. So, this is my answer. OP can always post an answer, too. – Gerry Myerson Mar 30 '12 at 1:26

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