We have indeterminateness of type
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \sqrt{x} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} \tan{\left(x \right)} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{\sqrt{x}}{\tan{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \sqrt{x}}{\frac{d}{d x} \tan{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{1}{2 \sqrt{x} \left(\tan^{2}{\left(x \right)} + 1\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{1}{2 \sqrt{x} \left(\tan^{2}{\left(x \right)} + 1\right)}\right)$$
=
$$\infty$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)