Answer
$0$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to +\infty} \dfrac{\ln (\ln x)}{\sqrt x}$
Apply L-Hospital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to +\infty} \dfrac{\ln (\ln x)}{\sqrt x} =\dfrac{1}{\frac{1}{2}} \lim\limits_{x \to +\infty} \dfrac{1}{x} \times \dfrac{1}{\ln (x) x^{-1/2}} $
Thus, we have: $\lim\limits_{x \to +\infty} \dfrac{\ln (\ln x)}{\sqrt x}=\dfrac{2}{ \infty}=0$
