# How do I take the limit as $n$ goes to $\infty$ of $\dfrac{\sqrt{n}}{\log(n)}$?

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

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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