Tenemos la indeterminación de tipo
oo/oo,
tal que el límite para el numerador es
$$\lim_{x \to \infty} \frac{1}{e^{\frac{1}{3 x + 1}} - 1} = \infty$$
y el límite para el denominador es
$$\lim_{x \to \infty} \frac{1}{\operatorname{asin}{\left(\frac{2}{x} \right)}} = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to \infty}\left(\frac{\operatorname{asin}{\left(\frac{2}{x} \right)}}{e^{\frac{1}{3 x + 1}} - 1}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \frac{1}{e^{\frac{1}{3 x + 1}} - 1}}{\frac{d}{d x} \frac{1}{\operatorname{asin}{\left(\frac{2}{x} \right)}}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{3 x^{2} \sqrt{1 - \frac{4}{x^{2}}} e^{\frac{1}{3 x + 1}} \operatorname{asin}^{2}{\left(\frac{2}{x} \right)}}{2 \left(3 x + 1\right)^{2} \left(e^{\frac{1}{3 x + 1}} - 1\right)^{2}}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{1}{\left(e^{\frac{2}{3 x + 1}} - 2 e^{\frac{1}{3 x + 1}} + 1\right) \left(\frac{6}{\operatorname{asin}^{2}{\left(\frac{2}{x} \right)}} + \frac{4}{x \operatorname{asin}^{2}{\left(\frac{2}{x} \right)}} + \frac{2}{3 x^{2} \operatorname{asin}^{2}{\left(\frac{2}{x} \right)}}\right)}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \frac{1}{e^{\frac{2}{3 x + 1}} - 2 e^{\frac{1}{3 x + 1}} + 1}}{\frac{d}{d x} \left(\frac{6}{\operatorname{asin}^{2}{\left(\frac{2}{x} \right)}} + \frac{4}{x \operatorname{asin}^{2}{\left(\frac{2}{x} \right)}} + \frac{2}{3 x^{2} \operatorname{asin}^{2}{\left(\frac{2}{x} \right)}}\right)}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{6 e^{\frac{2}{3 x + 1}}}{\left(3 x + 1\right)^{2}} - \frac{6 e^{\frac{1}{3 x + 1}}}{\left(3 x + 1\right)^{2}}}{\left(e^{\frac{2}{3 x + 1}} - 2 e^{\frac{1}{3 x + 1}} + 1\right)^{2} \left(- \frac{4}{x^{2} \operatorname{asin}^{2}{\left(\frac{2}{x} \right)}} + \frac{24}{x^{2} \sqrt{1 - \frac{4}{x^{2}}} \operatorname{asin}^{3}{\left(\frac{2}{x} \right)}} - \frac{4}{3 x^{3} \operatorname{asin}^{2}{\left(\frac{2}{x} \right)}} + \frac{16}{x^{3} \sqrt{1 - \frac{4}{x^{2}}} \operatorname{asin}^{3}{\left(\frac{2}{x} \right)}} + \frac{8}{3 x^{4} \sqrt{1 - \frac{4}{x^{2}}} \operatorname{asin}^{3}{\left(\frac{2}{x} \right)}}\right)}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{6 e^{\frac{2}{3 x + 1}}}{\left(3 x + 1\right)^{2}} - \frac{6 e^{\frac{1}{3 x + 1}}}{\left(3 x + 1\right)^{2}}}{\left(e^{\frac{2}{3 x + 1}} - 2 e^{\frac{1}{3 x + 1}} + 1\right)^{2} \left(- \frac{4}{x^{2} \operatorname{asin}^{2}{\left(\frac{2}{x} \right)}} + \frac{24}{x^{2} \sqrt{1 - \frac{4}{x^{2}}} \operatorname{asin}^{3}{\left(\frac{2}{x} \right)}} - \frac{4}{3 x^{3} \operatorname{asin}^{2}{\left(\frac{2}{x} \right)}} + \frac{16}{x^{3} \sqrt{1 - \frac{4}{x^{2}}} \operatorname{asin}^{3}{\left(\frac{2}{x} \right)}} + \frac{8}{3 x^{4} \sqrt{1 - \frac{4}{x^{2}}} \operatorname{asin}^{3}{\left(\frac{2}{x} \right)}}\right)}\right)$$
=
$$6$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 2 vez (veces)