Finding the limit of $F(x)=\frac{x^2-4}{|x+2|}$

Question

Let $F(x)=\dfrac{x^2-4}{|x+2|}$ and find the following limits

$(a) \; \; \lim_{x \to -2^-}F(x)=$
$(b) \; \; \lim_{x \to -2^+}F(x)=-4$
$(c) \; \; \lim_{x \to -2}F(x)=DNE$

I substituted $-2$ to find $(b)$ and I guessed on $(c)$. I don't know how to solve for $(a).$ How do I go about solving this? I'm also confused a little on how to graph this function.

I did factor it as: $F(x)=\dfrac{(x+2)(x-2)}{(x+2)}$ and then I canceled similar terms.

Thank you.

Answer 1

Factor the top as $(x+2)(x-2)$, and notice that as $x \rightarrow -2^{+}$, then the denominator becomes $-(x+2)$.

As $x \rightarrow -2^{-}$, then the denominator becomes $(x+2)$. Now do cancellations, and substitute $x=2$

Answer 2

HINT

$$F(x)= \left\{ \begin{array}{lr} \dfrac{x^2-4}{x+2} & : x \lt -2\\~\\ \dfrac{x^2-4}{-(x+2)} & : x \gt -2 \\ \end{array} \right.$$