Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 8|

Evaluate

\lim\limits_{x\to 8}\:\frac{\sqrt[3]{x}-2}{x-8}

 

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

\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}

 

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