If lim _{x rightarrow 0} frac{3^{x^3}-left(1- | vedclass

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(B) Given the limit $L = \lim _{x \rightarrow 0} \frac{3^{x^3}-\left(1-x^3\right)^{2 / 3}}{x^2 \sin x}$.Since $\sin x \approx x$ as $x \rightarrow 0$,

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$2$

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(B) Given the limit $L = \lim _{x ightarrow 0} \frac{3^{x^3}-\left(1-x^3ight)^{2 / 3}}{x^2 \sin x}$.Since $\sin x \approx x$ as $x ightarrow 0$, the expression becomes $\lim _{x ightarrow 0} \frac{3^{x^3}-\left(1-x^3ight)^{2 / 3}}{x^3}$.Using the series expansions $a^u = 1 + u \ln a + O(u^2)$ and $(1+u)^n = 1 + nu + O(u^2)$, we have:$3^{x^3} = 1 + x^3 \ln 3 + O(x^6)$$(1-x^3)^{2/3} = 1 - \frac{2}{3}x^3 + O(x^6)$Substituting these into the limit:$L = \lim _{x ightarrow 0} \frac{(1 + x^3 \ln 3) - (1 - \frac{2}{3}x^3)}{x^3} = \lim _{x ightarrow 0} \frac{x^3(\ln 3 + \frac{2}{3})}{x^3} = \ln 3 + \frac{2}{3} = \frac{2}{3} + \log_e 3$.Comparing this with $p + \log q$, we get $p = \frac{2}{3}$ and $q = 3$.Therefore, $pq = \frac{2}{3} 	imes 3 = 2$.

If $\lim _{x \rightarrow 0} \frac{3^{x^3}-\left(1-x^3\right)^{2 / 3}}{x^2 \sin x}=p+\log q$, then $pq=$

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