Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 1^+} \tan^{3}{\left(x^{2} - 1 \right)} = 0$$
y el límite para el denominador es
$$\lim_{x \to 1^+}\left(- \frac{2 \sqrt{x} e^{x}}{e \log{\left(x + 1 \right)}} + \frac{2 \sqrt{x}}{\log{\left(x + 1 \right)}} + \frac{x e^{x}}{e \log{\left(x + 1 \right)}} - \frac{x}{\log{\left(x + 1 \right)}} + \frac{e^{x}}{e \log{\left(x + 1 \right)}} - \frac{1}{\log{\left(x + 1 \right)}}\right) = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 1^+}\left(\frac{\log{\left(x + 1 \right)} \tan^{3}{\left(x^{2} - 1 \right)}}{\left(e^{x - 1} - 1\right) \left(\sqrt{x} - 1\right)^{2}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to 1^+}\left(\frac{\log{\left(x + 1 \right)} \tan^{3}{\left(x^{2} - 1 \right)}}{\left(\sqrt{x} - 1\right)^{2} \left(e^{x - 1} - 1\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\frac{d}{d x} \tan^{3}{\left(x^{2} - 1 \right)}}{\frac{d}{d x} \left(- \frac{2 \sqrt{x} e^{x}}{e \log{\left(x + 1 \right)}} + \frac{2 \sqrt{x}}{\log{\left(x + 1 \right)}} + \frac{x e^{x}}{e \log{\left(x + 1 \right)}} - \frac{x}{\log{\left(x + 1 \right)}} + \frac{e^{x}}{e \log{\left(x + 1 \right)}} - \frac{1}{\log{\left(x + 1 \right)}}\right)}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{6 x \left(\tan^{2}{\left(x^{2} - 1 \right)} + 1\right) \tan^{2}{\left(x^{2} - 1 \right)}}{- \frac{2 \sqrt{x} e^{x}}{e \log{\left(x + 1 \right)}} + \frac{2 \sqrt{x} e^{x}}{e \left(x + 1\right) \log{\left(x + 1 \right)}^{2}} - \frac{2 \sqrt{x}}{\left(x + 1\right) \log{\left(x + 1 \right)}^{2}} + \frac{x e^{x}}{e \log{\left(x + 1 \right)}} - \frac{x e^{x}}{e \left(x + 1\right) \log{\left(x + 1 \right)}^{2}} + \frac{x}{\left(x + 1\right) \log{\left(x + 1 \right)}^{2}} + \frac{2 e^{x}}{e \log{\left(x + 1 \right)}} - \frac{1}{\log{\left(x + 1 \right)}} - \frac{e^{x}}{e \left(x + 1\right) \log{\left(x + 1 \right)}^{2}} + \frac{1}{\left(x + 1\right) \log{\left(x + 1 \right)}^{2}} - \frac{e^{x}}{e \sqrt{x} \log{\left(x + 1 \right)}} + \frac{1}{\sqrt{x} \log{\left(x + 1 \right)}}}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{6 \tan^{2}{\left(x^{2} - 1 \right)}}{- \frac{2 \sqrt{x} e^{x}}{e \log{\left(x + 1 \right)}} + \frac{2 \sqrt{x} e^{x}}{e \left(x + 1\right) \log{\left(x + 1 \right)}^{2}} - \frac{2 \sqrt{x}}{\left(x + 1\right) \log{\left(x + 1 \right)}^{2}} + \frac{x e^{x}}{e \log{\left(x + 1 \right)}} - \frac{x e^{x}}{e \left(x + 1\right) \log{\left(x + 1 \right)}^{2}} + \frac{x}{\left(x + 1\right) \log{\left(x + 1 \right)}^{2}} + \frac{2 e^{x}}{e \log{\left(x + 1 \right)}} - \frac{1}{\log{\left(x + 1 \right)}} - \frac{e^{x}}{e \left(x + 1\right) \log{\left(x + 1 \right)}^{2}} + \frac{1}{\left(x + 1\right) \log{\left(x + 1 \right)}^{2}} - \frac{e^{x}}{e \sqrt{x} \log{\left(x + 1 \right)}} + \frac{1}{\sqrt{x} \log{\left(x + 1 \right)}}}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{6 \tan^{2}{\left(x^{2} - 1 \right)}}{- \frac{2 \sqrt{x} e^{x}}{e \log{\left(x + 1 \right)}} + \frac{2 \sqrt{x} e^{x}}{e \left(x + 1\right) \log{\left(x + 1 \right)}^{2}} - \frac{2 \sqrt{x}}{\left(x + 1\right) \log{\left(x + 1 \right)}^{2}} + \frac{x e^{x}}{e \log{\left(x + 1 \right)}} - \frac{x e^{x}}{e \left(x + 1\right) \log{\left(x + 1 \right)}^{2}} + \frac{x}{\left(x + 1\right) \log{\left(x + 1 \right)}^{2}} + \frac{2 e^{x}}{e \log{\left(x + 1 \right)}} - \frac{1}{\log{\left(x + 1 \right)}} - \frac{e^{x}}{e \left(x + 1\right) \log{\left(x + 1 \right)}^{2}} + \frac{1}{\left(x + 1\right) \log{\left(x + 1 \right)}^{2}} - \frac{e^{x}}{e \sqrt{x} \log{\left(x + 1 \right)}} + \frac{1}{\sqrt{x} \log{\left(x + 1 \right)}}}\right)$$
=
$$32 \log{\left(2 \right)}$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 1 vez (veces)