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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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    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 $c$.2012-03-30

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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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    Ops. I've just notices you used $O(x^3)$ expansion of $e^x.$ Ignore my previous comment.2012-03-30
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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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    The question was, "should I use l'Hopital's Rule?" I could have si$m$ply 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 ga$m$e away myself. So, this is my answer. OP can always post an answer, too.2012-03-30