We have indeterminateness of type
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+}\left(- x \cot{\left(x \right)} + 1\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+} x^{2} = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(- \frac{\cot{\left(x \right)}}{x} + \frac{1}{x^{2}}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 0^+}\left(\frac{- x \cot{\left(x \right)} + 1}{x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- x \cot{\left(x \right)} + 1\right)}{\frac{d}{d x} x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x \cot^{2}{\left(x \right)} + x - \cot{\left(x \right)}}{2 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(x \cot^{2}{\left(x \right)} + x - \cot{\left(x \right)}\right)}{\frac{d}{d x} 2 x}\right)$$
=
$$\lim_{x \to 0^+}\left(- x \cot^{3}{\left(x \right)} - x \cot{\left(x \right)} + \cot^{2}{\left(x \right)} + 1\right)$$
=
$$\lim_{x \to 0^+}\left(- x \cot^{3}{\left(x \right)} - x \cot{\left(x \right)} + \cot^{2}{\left(x \right)} + 1\right)$$
=
$$\frac{1}{3}$$
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)