Limit Evaluation of Exponential and Algebraic Functions

To find the value of $6\beta$, we first evaluate the given limit: $\beta = \lim_{x \to 0} \frac{e^{x^3} - (1 - x^3)^{\frac{1}{3}} + \left((1 - x^2)^{\frac{1}{2}} - 1\right) \sin x}{x \sin^2 x}$

We use the Taylor series expansions near $x = 0$ for each component of the expression up to the $x^3$ term:

1. Expansion of $e^{x^3}$: $e^{x^3} = 1 + x^3 + O(x^6)$

2. Expansion of $(1 - x^3)^{\frac{1}{3}}$: $(1 - x^3)^{\frac{1}{3}} = 1 - \frac{1}{3}x^3 + O(x^6)$

3. Expansion of $\left((1 - x^2)^{\frac{1}{2}} - 1\right) \sin x$: $(1 - x^2)^{\frac{1}{2}} = 1 - \frac{1}{2}x^2 + O(x^4)$ $(1 - x^2)^{\frac{1}{2}} - 1 = -\frac{1}{2}x^2 + O(x^4)$ Since $\sin x = x - \frac{x^3}{6} + O(x^5)$, we have: $\left((1 - x^2)^{\frac{1}{2}} - 1\right) \sin x = \left(-\frac{1}{2}x^2 + O(x^4)\right)\left(x + O(x^3)\right) = -\frac{1}{2}x^3 + O(x^5)$

Now, substitute these expansions into the numerator: $\text{Numerator} = \left(1 + x^3\right) - \left(1 - \frac{1}{3}x^3\right) - \frac{1}{2}x^3 + O(x^5)$ $\text{Numerator} = \left(1 - 1\right) + \left(1 + \frac{1}{3} - \frac{1}{2}\right)x^3 + O(x^5)$ $\text{Numerator} = \left(\frac{6 + 2 - 3}{6}\right)x^3 + O(x^5) = \frac{5}{6}x^3 + O(x^5)$

For the denominator, as $x \to 0$: $x \sin^2 x = x \left(x + O(x^3)\right)^2 = x^3 + O(x^5)$

Substituting the numerator and denominator back into the limit: $\beta = \lim_{x \to 0} \frac{\frac{5}{6}x^3 + O(x^5)}{x^3 + O(x^5)} = \lim_{x \to 0} \frac{\frac{5}{6} + O(x^2)}{1 + O(x^2)} = \frac{5}{6}$

Thus, the value of $6\beta$ is: $6\beta = 6 \times \frac{5}{6} = 5$