Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{n \to 0^+} \sqrt{- \sqrt{n} + 2 n^{2}} = 0$$
y el límite para el denominador es
$$\lim_{n \to 0^+}\left(n + \sqrt{n^{3}}\right) = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{n \to 0^+}\left(\frac{\sqrt{- \sqrt{n} + 2 n^{2}}}{n + \sqrt{n^{3}}}\right)$$
=
$$\lim_{n \to 0^+}\left(\frac{\frac{d}{d n} \sqrt{- \sqrt{n} + 2 n^{2}}}{\frac{d}{d n} \left(n + \sqrt{n^{3}}\right)}\right)$$
=
$$\lim_{n \to 0^+}\left(\frac{2 n - \frac{1}{4 \sqrt{n}}}{\left(1 + \frac{3 \sqrt{n^{3}}}{2 n}\right) \sqrt{- \sqrt{n} + 2 n^{2}}}\right)$$
=
$$\lim_{n \to 0^+}\left(\frac{\frac{d}{d n} \frac{1}{1 + \frac{3 \sqrt{n^{3}}}{2 n}}}{\frac{d}{d n} \frac{\sqrt{- \sqrt{n} + 2 n^{2}}}{2 n - \frac{1}{4 \sqrt{n}}}}\right)$$
=
$$\lim_{n \to 0^+}\left(- \frac{3 \sqrt{n^{3}}}{4 n^{2} \left(1 + \frac{3 \sqrt{n^{3}}}{2 n}\right)^{2} \left(\frac{\left(-2 - \frac{1}{8 n^{\frac{3}{2}}}\right) \sqrt{- \sqrt{n} + 2 n^{2}}}{\left(2 n - \frac{1}{4 \sqrt{n}}\right)^{2}} + \frac{1}{\sqrt{- \sqrt{n} + 2 n^{2}}}\right)}\right)$$
=
$$\lim_{n \to 0^+}\left(- \frac{3 \sqrt{n^{3}}}{4 n^{2} \left(- \frac{\sqrt{- \sqrt{n} + 2 n^{2}}}{32 n^{\frac{7}{2}} + \frac{\sqrt{n}}{2} - 8 n^{2}} - \frac{2 \sqrt{- \sqrt{n} + 2 n^{2}}}{- \sqrt{n} + 4 n^{2} + \frac{1}{16 n}} + \frac{1}{\sqrt{- \sqrt{n} + 2 n^{2}}}\right)}\right)$$
=
$$\lim_{n \to 0^+}\left(\frac{\frac{d}{d n} \frac{1}{- \frac{\sqrt{- \sqrt{n} + 2 n^{2}}}{32 n^{\frac{7}{2}} + \frac{\sqrt{n}}{2} - 8 n^{2}} - \frac{2 \sqrt{- \sqrt{n} + 2 n^{2}}}{- \sqrt{n} + 4 n^{2} + \frac{1}{16 n}} + \frac{1}{\sqrt{- \sqrt{n} + 2 n^{2}}}}}{\frac{d}{d n} \left(- \frac{4 n^{2}}{3 \sqrt{n^{3}}}\right)}\right)$$
=
$$\lim_{n \to 0^+}\left(- \frac{3 \left(\frac{\sqrt{- \sqrt{n} + 2 n^{2}} \left(- 112 n^{\frac{5}{2}} + 16 n - \frac{1}{4 \sqrt{n}}\right)}{\left(32 n^{\frac{7}{2}} + \frac{\sqrt{n}}{2} - 8 n^{2}\right)^{2}} + \frac{2 \sqrt{- \sqrt{n} + 2 n^{2}} \left(- 8 n + \frac{1}{16 n^{2}} + \frac{1}{2 \sqrt{n}}\right)}{\left(- \sqrt{n} + 4 n^{2} + \frac{1}{16 n}\right)^{2}} + \frac{2 n - \frac{1}{4 \sqrt{n}}}{\sqrt{- \sqrt{n} + 2 n^{2}} \left(32 n^{\frac{7}{2}} + \frac{\sqrt{n}}{2} - 8 n^{2}\right)} + \frac{2 \left(2 n - \frac{1}{4 \sqrt{n}}\right)}{\sqrt{- \sqrt{n} + 2 n^{2}} \left(- \sqrt{n} + 4 n^{2} + \frac{1}{16 n}\right)} - \frac{- 2 n + \frac{1}{4 \sqrt{n}}}{\left(- \sqrt{n} + 2 n^{2}\right)^{\frac{3}{2}}}\right) \sqrt{n^{3}}}{2 n \left(- \frac{\sqrt{- \sqrt{n} + 2 n^{2}}}{32 n^{\frac{7}{2}} + \frac{\sqrt{n}}{2} - 8 n^{2}} - \frac{2 \sqrt{- \sqrt{n} + 2 n^{2}}}{- \sqrt{n} + 4 n^{2} + \frac{1}{16 n}} + \frac{1}{\sqrt{- \sqrt{n} + 2 n^{2}}}\right)^{2}}\right)$$
=
$$\lim_{n \to 0^+}\left(- \frac{3 \left(\frac{\sqrt{- \sqrt{n} + 2 n^{2}} \left(- 112 n^{\frac{5}{2}} + 16 n - \frac{1}{4 \sqrt{n}}\right)}{\left(32 n^{\frac{7}{2}} + \frac{\sqrt{n}}{2} - 8 n^{2}\right)^{2}} + \frac{2 \sqrt{- \sqrt{n} + 2 n^{2}} \left(- 8 n + \frac{1}{16 n^{2}} + \frac{1}{2 \sqrt{n}}\right)}{\left(- \sqrt{n} + 4 n^{2} + \frac{1}{16 n}\right)^{2}} + \frac{2 n - \frac{1}{4 \sqrt{n}}}{\sqrt{- \sqrt{n} + 2 n^{2}} \left(32 n^{\frac{7}{2}} + \frac{\sqrt{n}}{2} - 8 n^{2}\right)} + \frac{2 \left(2 n - \frac{1}{4 \sqrt{n}}\right)}{\sqrt{- \sqrt{n} + 2 n^{2}} \left(- \sqrt{n} + 4 n^{2} + \frac{1}{16 n}\right)} - \frac{- 2 n + \frac{1}{4 \sqrt{n}}}{\left(- \sqrt{n} + 2 n^{2}\right)^{\frac{3}{2}}}\right) \sqrt{n^{3}}}{2 n \left(- \frac{\sqrt{- \sqrt{n} + 2 n^{2}}}{32 n^{\frac{7}{2}} + \frac{\sqrt{n}}{2} - 8 n^{2}} - \frac{2 \sqrt{- \sqrt{n} + 2 n^{2}}}{- \sqrt{n} + 4 n^{2} + \frac{1}{16 n}} + \frac{1}{\sqrt{- \sqrt{n} + 2 n^{2}}}\right)^{2}}\right)$$
=
$$\infty i$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 3 vez (veces)