# lim x--> ?

Jan 25, 2017

See the explanation below

#### Explanation:

You can simplify the expression of the function noting that:

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

So that we have:

$\left\{\begin{matrix}f \left(x\right) = x - 4 \text{ if " x < 0 \\ f(x) = x+4 " if } x > 0\end{matrix}\right.$

We have then:

(a) ${\lim}_{x \to - 4} f \left(x\right) = {\lim}_{x \to - 4} \left(x - 4\right) = - 8$

(b) ${\lim}_{x \to 0} f \left(x\right)$ does not exist, since:

${\lim}_{x \to {0}^{-}} f \left(x\right) = {\lim}_{x \to {0}^{-}} \left(x - 4\right) = - 4$

${\lim}_{x \to {0}^{+}} f \left(x\right) = {\lim}_{x \to {0}^{+}} \left(x + 4\right) = 4$

(c) ${\lim}_{x \to 4} f \left(x\right) = {\lim}_{x \to 4} \left(x + 4\right) = 8$