Differential and Integral Calculus by Feliciano and Uy: Limit of a Function in Indeterminate Form, Exercise 1.3, Problem 1

Evaluate

\lim\limits_{x\to 4}\left(\frac{x^3-64}{x^2-16}\right)

SOLUTION:

A straight substitution of x=4 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 4}\left(\frac{x^3-64}{x^2-16}\right)=\lim\limits_{x\to 4}\left(\frac{\left(x-4\right)\left(x^2+4x+16\right)}{\left(x+4\right)\left(x-4\right)}\right)

=\lim\limits_{x\to 4}\left(\frac{x^2+4x+16}{x+4}\right)

=\frac{\left(4\right)^2+4\left(4\right)+16}{4+4}

=\frac{48}{8}

=6

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