Respuesta :
Expand the limand as
[tex]\dfrac{e^{x^3} - (1 - x^3)^{1/3} + \left((1-x^2)^{1/2} - 1\right) \sin(x)}{x \sin^2(x)} \\\\ ~~~~~~~~ = \dfrac{e^{x^3} - (1-x^3)^{1/3}}{x^3} \cdot \dfrac{x^2}{\sin^2(x)} + \dfrac{(1-x^2)^{1/2} - 1}{x^2} \cdot \dfrac{x}{\sin(x)}[/tex]
Recall that for [tex]a\neq0[/tex],
[tex]\displaystyle \lim_{x\to0} \frac{\sin(ax)}{ax} = 1[/tex]
Then
[tex]\displaystyle \beta = \lim_{x\to0} \left( \dfrac{e^{x^3} - (1-x^3)^{1/3}}{x^3} + \dfrac{(1-x^2)^{1/2} - 1}{x^2}\right)[/tex]
Rationalize the numerators.
[tex]\dfrac{e^{x^3} - (1-x^3)^{1/3}}{x^3} = \dfrac{e^{3x^3} - (1 - x^3)}{x^3 \left(e^{2x^3} + e^{x^3} (1-x^3)^{1/3} + (1-x^3)^{2/3}\right)} \\\\ ~~~~~~~~~~~ = \left(\dfrac{e^{3x^3} - 1}{x^3} + 1\right) \cdot\dfrac1{e^{2x^3} + e^{x^3} (1-x^3)^{1/3} + (1-x^3)^{2/3}}[/tex]
and observe that
[tex]\dfrac1{e^{2x^3} + e^{x^3} (1-x^3)^{1/3} + (1-x^3)^{2/3}} \to \dfrac13[/tex]
[tex]\dfrac{(1-x^2)^{1/2}-1}{x^2} = \dfrac{(1-x^2) - 1}{x^2 \left((1-x^2)^{1/2} + 1\right)} \\\\ ~~~~~~~~ = -\dfrac1{(1-x^2)^{1/2} + 1} \to -\dfrac12[/tex]
This leaves us with
[tex]\displaystyle \beta = \lim_{x\to0} \frac{e^{3x^3} - 1}{3x^3} - \frac16[/tex]
The remaining limit is the derivative of [tex]e^{3x^3}[/tex] at [tex]x=0[/tex], so its value is 1, and we find
[tex]\beta = 1 - \dfrac16 = \dfrac56 \implies 6\beta = \boxed{5}[/tex]