The value of $$\mathop {\lim }\limits_{{\text{x}} \to 8} \frac{{{{\text{x}}^{\frac{1}{3}}} – 2}}{{\left( {{\text{x}} – 8} \right)}}$$ A. $$\frac{1}{{16}}$$ B. $$\frac{1}{{12}}$$ C. $$\frac{1}{8}$$ D. $$\frac{1}{4}$$

The correct answer is $\boxed{\frac{1}{8}}$.

To find the limit, we can substitute $x=8$ into the expression. However, this results in the indeterminate form $\frac{0}{0}$. This means that we need to use a different method to find the limit.

One way to do this is to use l’Hôpital’s rule. L’Hôpital’s rule states that if $\lim_{x\to a}\frac{f(x)}{g(x)}$ is equal to the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x\to a}\frac{f'(x)}{g'(x)}$ exists and is equal to the limit of $\frac{f(x)}{g(x)}$.

In this case, we have

$$\begin{align}
\lim_{x\to 8} \frac{{{{\text{x}}^{\frac{1}{3}}} – 2}}{{\left( {{\text{x}} – 8} \right)}} &= \lim_{x\to 8} \frac{\frac{1}{3}{{x}^{^{ – \frac{2}{3}}}}}{{1}} \
&= \lim_{x\to 8} \frac{1}{3}{{x}^{^{ – \frac{2}{3} – 1}}} \
&= \lim_{x\to 8} \frac{1}{3}{{x}^{^{ – \frac{5}{3}}}} \
&= \frac{1}{3} \cdot \frac{1}{{8}^{^{ – \frac{5}{3}}}} \
&= \frac{1}{3} \cdot \frac{1}{{2^5}} \
&= \frac{1}{3} \cdot \frac{1}{32} \
&= \frac{1}{8}.
\end{align}$$

Therefore, the value of the limit is $\boxed{\frac{1}{8}}$.