Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 8

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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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