Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 1

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PROBLEM:

Evaluate lim⁡x→4(x3−64x2−16)\displaystyle \lim\limits_{x\to 4}\left(\frac{x^3-64}{x^2-16}\right)

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SOLUTION:

A straight substitution of x=4 x=4 leads to the indeterminate form 00 \frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→4(x3−64x2−16)=lim⁡x→4((x−4)(x2+4x+16)(x+4)(x−4))=lim⁡x→4(x2+4x+16x+4)=(4)2+4(4)+164+4=488=6  (Answer)\begin{align*} \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 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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