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PROBLEM:
Evaluate \displaystyle \lim\limits_{x\to 8}\:\frac{\sqrt[3]{x}-2}{x-8}.
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SOLUTION:
A straight substitution of x=8 leads to the indeterminate form \frac{0}{0} which is meaningless.
Therefore, to evaluate the limit of the given function, we proceed as follows
\begin{align*}
\\
\lim\limits_{x\to 8}\:\frac{\sqrt[3]{x}-2}{x-8}& =\lim\limits_{x\to \:8}\:\frac{\sqrt[3]{x}-2}{x-8}\cdot \frac{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}\\
\\
& =\lim\limits_{x\to 8}\:\frac{x-8}{\left(x-8\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}\\
\\
& =\lim\limits_{x\to 8}\:\frac{1}{\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}\\
\\
& =\frac{1}{\left(\sqrt[3]{8^2}+2\sqrt[3]{8}+4\right)}\\
\\
& =\frac{1}{4+4+4}\\
\\
& =\frac{1}{12} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\\
\end{align*}
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