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

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

SOLUTION

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