We have indeterminateness of type
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+}\left(x - \operatorname{atan}{\left(x \right)}\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} x^{4} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{x - \operatorname{atan}{\left(x \right)}}{x^{4}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(x - \operatorname{atan}{\left(x \right)}\right)}{\frac{d}{d x} x^{4}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{1 - \frac{1}{x^{2} + 1}}{4 x^{3}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(1 - \frac{1}{x^{2} + 1}\right)}{\frac{d}{d x} 4 x^{3}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{1}{6 x \left(x^{2} + 1\right)^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{1}{6 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{1}{6 x}\right)$$
=
$$\infty$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 2 time(s)