Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+}\left(- \sqrt{\sin{\left(x \right)}} + \sqrt{\tan{\left(x \right)}}\right) = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+} x^{\frac{5}{2}} = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\frac{- \sqrt{\sin{\left(x \right)}} + \sqrt{\tan{\left(x \right)}}}{x^{\frac{5}{2}}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- \sqrt{\sin{\left(x \right)}} + \sqrt{\tan{\left(x \right)}}\right)}{\frac{d}{d x} x^{\frac{5}{2}}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 \left(\frac{\tan^{\frac{3}{2}}{\left(x \right)}}{2} + \frac{1}{2 \sqrt{\tan{\left(x \right)}}} - \frac{\cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}}\right)}{5 x^{\frac{3}{2}}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(\frac{\tan^{\frac{3}{2}}{\left(x \right)}}{2} + \frac{1}{2 \sqrt{\tan{\left(x \right)}}} - \frac{\cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}}\right)}{\frac{d}{d x} \frac{5 x^{\frac{3}{2}}}{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{4 \left(\frac{\sqrt{\sin{\left(x \right)}}}{2} + \frac{3 \tan^{\frac{5}{2}}{\left(x \right)}}{4} + \frac{\sqrt{\tan{\left(x \right)}}}{2} - \frac{1}{4 \tan^{\frac{3}{2}}{\left(x \right)}} + \frac{\cos^{2}{\left(x \right)}}{4 \sin^{\frac{3}{2}}{\left(x \right)}}\right)}{15 \sqrt{x}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(\frac{\sqrt{\sin{\left(x \right)}}}{2} + \frac{3 \tan^{\frac{5}{2}}{\left(x \right)}}{4} + \frac{\sqrt{\tan{\left(x \right)}}}{2} - \frac{1}{4 \tan^{\frac{3}{2}}{\left(x \right)}} + \frac{\cos^{2}{\left(x \right)}}{4 \sin^{\frac{3}{2}}{\left(x \right)}}\right)}{\frac{d}{d x} \frac{15 \sqrt{x}}{4}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{8 \sqrt{x} \left(- \frac{- \frac{3 \tan^{2}{\left(x \right)}}{2} - \frac{3}{2}}{4 \tan^{\frac{5}{2}}{\left(x \right)}} + \frac{\frac{\tan^{2}{\left(x \right)}}{2} + \frac{1}{2}}{2 \sqrt{\tan{\left(x \right)}}} + \frac{3 \left(\frac{5 \tan^{2}{\left(x \right)}}{2} + \frac{5}{2}\right) \tan^{\frac{3}{2}}{\left(x \right)}}{4} - \frac{\cos{\left(x \right)}}{4 \sqrt{\sin{\left(x \right)}}} - \frac{3 \cos^{3}{\left(x \right)}}{8 \sin^{\frac{5}{2}}{\left(x \right)}}\right)}{15}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{8 \sqrt{x} \left(- \frac{- \frac{3 \tan^{2}{\left(x \right)}}{2} - \frac{3}{2}}{4 \tan^{\frac{5}{2}}{\left(x \right)}} + \frac{\frac{\tan^{2}{\left(x \right)}}{2} + \frac{1}{2}}{2 \sqrt{\tan{\left(x \right)}}} + \frac{3 \left(\frac{5 \tan^{2}{\left(x \right)}}{2} + \frac{5}{2}\right) \tan^{\frac{3}{2}}{\left(x \right)}}{4} - \frac{\cos{\left(x \right)}}{4 \sqrt{\sin{\left(x \right)}}} - \frac{3 \cos^{3}{\left(x \right)}}{8 \sin^{\frac{5}{2}}{\left(x \right)}}\right)}{15}\right)$$
=
$$\frac{1}{4}$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 3 vez (veces)