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(B) Given the limit $\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)$.As $x \rightarrow 2^{+}$,$x$ is slightly grea

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$\frac{8}{3}$

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(B) Given the limit $\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)$.As $x \rightarrow 2^{+}$,$x$ is slightly greater than $2$,so $[x] = 2$.Also,as $x \rightarrow 2^{+}$,$\frac{x}{3}$ is slightly greater than $\frac{2}{3}$,so $\left[\frac{x}{3}\right] = 0$.Substituting these values into the expression:$\lim _{x}$ ${\rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right) = \frac{2^3}{3} - 0^3 = \frac{8}{3} - 0 = \frac{8}{3}$.

Let $[x]$ denote the greatest integer less than or equal to $x$. Then $\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)=$

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