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

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

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

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

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

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

$=\frac{48}{8}$

$=6$