Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 1^+}\left(\sqrt{x} - 1\right) = 0$$
y el límite para el denominador es
$$\lim_{x \to 1^+} \frac{1}{\cot{\left(\pi x \right)}} = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 1^+}\left(\left(\sqrt{x} - 1\right) \cot{\left(\pi x \right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\frac{d}{d x} \left(\sqrt{x} - 1\right)}{\frac{d}{d x} \frac{1}{\cot{\left(\pi x \right)}}}\right)$$
=
$$\lim_{x \to 1^+}\left(- \frac{\cot^{2}{\left(\pi x \right)}}{2 \pi \sqrt{x} \left(- \cot^{2}{\left(\pi x \right)} - 1\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(- \frac{\cot^{2}{\left(\pi x \right)}}{2 \pi \left(- \cot^{2}{\left(\pi x \right)} - 1\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\frac{d}{d x} \frac{1}{- \cot^{2}{\left(\pi x \right)} - 1}}{\frac{d}{d x} \left(- \frac{2 \pi}{\cot^{2}{\left(\pi x \right)}}\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\cot^{4}{\left(\pi x \right)}}{2 \pi \left(\cot^{4}{\left(\pi x \right)} + 2 \cot^{2}{\left(\pi x \right)} + 1\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\frac{d}{d x} \frac{1}{\cot^{4}{\left(\pi x \right)} + 2 \cot^{2}{\left(\pi x \right)} + 1}}{\frac{d}{d x} \frac{2 \pi}{\cot^{4}{\left(\pi x \right)}}}\right)$$
=
$$\lim_{x \to 1^+}\left(- \frac{\left(- 4 \pi \left(- \cot^{2}{\left(\pi x \right)} - 1\right) \cot^{3}{\left(\pi x \right)} - 4 \pi \left(- \cot^{2}{\left(\pi x \right)} - 1\right) \cot{\left(\pi x \right)}\right) \cot^{5}{\left(\pi x \right)}}{8 \pi^{2} \left(- \cot^{2}{\left(\pi x \right)} - 1\right) \left(\cot^{4}{\left(\pi x \right)} + 2 \cot^{2}{\left(\pi x \right)} + 1\right)^{2}}\right)$$
=
$$\lim_{x \to 1^+}\left(- \frac{\left(- 4 \pi \left(- \cot^{2}{\left(\pi x \right)} - 1\right) \cot^{3}{\left(\pi x \right)} - 4 \pi \left(- \cot^{2}{\left(\pi x \right)} - 1\right) \cot{\left(\pi x \right)}\right) \cot^{5}{\left(\pi x \right)}}{8 \pi^{2} \left(- \cot^{2}{\left(\pi x \right)} - 1\right) \left(\cot^{4}{\left(\pi x \right)} + 2 \cot^{2}{\left(\pi x \right)} + 1\right)^{2}}\right)$$
=
$$\frac{1}{2 \pi}$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 3 vez (veces)