Answer
The limit is not defined.
Work Step by Step
Our aim is to compute the value of $\lim\limits_{x \to \ 4} \dfrac{x^2+8}{x^2-2x-8}$.
Here, we need to use the formula of:
$\lim \limits_{x\to a}f(x)=f(a)$, this implies that:
$\lim\limits_{x \to \ 4} \dfrac{x^2+8}{x^2-2x-8}=\dfrac{(4)^2+8}{(4)^2-(2)(4)-8}\\=\dfrac{24}{0}$
Now, $\lim\limits_{x \to \ 4^{+}} \dfrac{x^2+8}{x^2-2x-8}=\dfrac{(4.01)^2+8}{(4.01)^2-(2)(4.01)-8}\\=+\infty$ and $\lim\limits_{x \to \ 4^{-}} \dfrac{x^2+8}{x^2-2x-8}=\dfrac{(3.99)^2+8}{(3.99)^2-(2)(3.99)-8}\\=-\infty$
This implies that $\lim \limits_{x\to e^{-}}f(x) \ne \lim \limits_{x\to e^{+}}f(x)$
Therefore, we can state that the limit is not defined.
