If beta = lim_{x rightarrow 0} frac{e^{x^3} - | vedclass

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(A) We are given $\beta = \lim_{x \rightarrow 0} \frac{e^{x^3} - (1 - x^3)^{1/3} + ((1 - x^2)^{1/2} - 1) \sin x}{x \sin^2 x}$.Using the standard limit

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$5$

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(A) We are given $\beta = \lim_{x ightarrow 0} \frac{e^{x^3} - (1 - x^3)^{1/3} + ((1 - x^2)^{1/2} - 1) \sin x}{x \sin^2 x}$.Using the standard limit $\lim_{x ightarrow 0} \frac{\sin x}{x} = 1$,we can rewrite the denominator as $x \sin^2 x = x^3 \left(\frac{\sin x}{x}ight)^2 \approx x^3$ as $x ightarrow 0$.Expanding the terms using Taylor series:$e^{x^3} = 1 + x^3 + O(x^6)$$(1 - x^3)^{1/3} = 1 - \frac{1}{3}x^3 + O(x^6)$$(1 - x^2)^{1/2} - 1 = (1 - \frac{1}{2}x^2 + O(x^4)) - 1 = -\frac{1}{2}x^2 + O(x^4)$Substituting these into the expression:$\beta = \lim_{x ightarrow 0} \frac{(1 + x^3) - (1 - \frac{1}{3}x^3) + (-\frac{1}{2}x^2) \cdot x}{x^3}$$\beta = \lim_{x ightarrow 0} \frac{1 + x^3 - 1 + \frac{1}{3}x^3 - \frac{1}{2}x^3}{x^3}$$\beta = \lim_{x}$ ${ightarrow 0} \frac{(1 + \frac{1}{3} - \frac{1}{2})x^3}{x^3} = 1 + \frac{1}{3} - \frac{1}{2} = \frac{6 + 2 - 3}{6} = \frac{5}{6}$.Therefore,$6 \beta = 6 	imes \frac{5}{6} = 5$.

If $\beta = \lim_{x \rightarrow 0} \frac{e^{x^3} - (1 - x^3)^{1/3} + ((1 - x^2)^{1/2} - 1) \sin x}{x \sin^2 x}$,then the value of $6 \beta$ is

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