Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 10|

Differential and Integral Calculus by Feliciano and Uy Solution Manual by


\lim\limits_{x\to 2}\left(\frac{x^3-8}{x^2-4\:}\right)


A straight substitution of  x=2 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 2}\left(\frac{x^3-8}{x^2-4\:}\right)=\lim\limits_{x\to 2}\left(\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x-2\right)}\right)

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

=\frac{2^2+2\cdot 2+4}{2+2}