# How do you find the limit of (lnx)^2/sqrtx as x->oo?

Feb 14, 2017

${\lim}_{x \to \infty} {\left(\ln x\right)}^{2} / \sqrt{x} = 0$

#### Explanation:

The limit:

${\lim}_{x \to \infty} {\left(\ln x\right)}^{2} / \sqrt{x}$

is in the indeterminate form: $\frac{\infty}{\infty}$ so we can use l'Hospital's rule:

${\lim}_{x \to \infty} {\left(\ln x\right)}^{2} / \sqrt{x} = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} {\left(\ln x\right)}^{2}}{\frac{d}{\mathrm{dx}} \sqrt{x}} = {\lim}_{x \to \infty} \frac{2 \ln x}{x} \frac{1}{\frac{1}{2} {x}^{\frac{3}{2}}} = {\lim}_{x \to \infty} \frac{4 \ln x}{x} ^ \left(\frac{5}{2}\right)$

and again:

${\lim}_{x \to \infty} \frac{4 \ln x}{x} ^ \left(\frac{5}{2}\right) = 4 {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln x}{\frac{d}{\mathrm{dx}} {x}^{\frac{5}{2}}} = 4 {\lim}_{x \to \infty} \frac{1}{x} \frac{1}{\frac{5}{2} {x}^{\frac{7}{2}}} = \frac{8}{5} {\lim}_{x \to \infty} \frac{1}{{x}^{\frac{9}{2}}} = 0$

In general we have that for every $\alpha > 0$:

${\lim}_{x \to \infty} \ln \frac{x}{x} ^ \alpha = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln x}{\frac{d}{\mathrm{dx}} {x}^{\alpha}} = {\lim}_{x \to \infty} \frac{1}{x} \frac{1}{\alpha {x}^{\alpha - 1}} = \frac{1}{\alpha} {\lim}_{x \to \infty} \frac{1}{x} ^ \alpha = 0$

so: $\ln x = o \left({x}^{\alpha}\right)$

So for any $a , b > 0$:

${\lim}_{x \to \infty} {\left(\ln x\right)}^{a} / {x}^{b} = {\lim}_{x \to \infty} {\left(\ln \frac{x}{x} ^ \left(\frac{b}{a}\right)\right)}^{a} = {\left({\lim}_{x \to \infty} \ln \frac{x}{x} ^ \left(\frac{b}{a}\right)\right)}^{a} = 0$