A limit with no l'Hospital's rule - $\lim_{n \to \infty} \frac{\ln n}{\sqrt{n}}$ [duplicate]

Solution 1:

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 $$\lim_{n \rightarrow \infty} \frac{\sqrt{n}}{\log(n)} = \lim_{x \rightarrow \infty} \frac{\exp(x/2)}{x} \geq \lim_{x \rightarrow \infty} \frac{\frac{x^2}{8}}{x} = \lim_{x \rightarrow \infty} \frac{x}{8} = \infty$$

Solution 2:

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.

Solution 3:

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