Answer
$+\infty$
Work Step by Step
The limit is:
$$\lim\limits_{x \to \infty}\frac{\sqrt x}{\ln(\ln x)}=\frac{\infty}{\infty}$$
Using the l'Hospital's rule it follows:
$$\lim\limits_{x \to \infty}\frac{\sqrt x}{\ln(\ln x)}=\lim\limits_{x \to \infty}\frac{\frac{1}{2\sqrt x}}{(\ln x)'\cdot \frac{1}{\ln x}}=\lim\limits_{x \to \infty} \frac{\frac{1}{2\sqrt x}}{\frac{1}{x}\cdot \frac{1}{\ln x}}=\lim\limits_{x \to \infty} \frac{x\ln x}{2\sqrt x}=\lim\limits_{x \to \infty} \frac{\sqrt x\ln x}{2}=+\infty$$
