Answer
See explanations.
Work Step by Step
Step 1. We only need to use the first equation of the piece-wise function as when $x\to-2, x\ne-2$. Thus we can identify the quantities as: $f(x)=x^2,c=-2,L=4$.
Step 2. For a given small value $\epsilon\gt0$, we need to find a value $\delta\gt0$ so that for all $x$ when $0\lt |x+2|\lt\delta$, we have $|x^2−4|\lt\epsilon$
Step 3. The last inequality gives $−\epsilon\lt x^2−4\lt\epsilon$ or $4−\epsilon\lt x^2\lt4+\epsilon$
Step 4. Assume $\epsilon\lt4$, and $x\lt0$; we can get $\sqrt {4−\epsilon}\lt -x \lt \sqrt {4+\epsilon}$ or $-\sqrt {4+\epsilon} \lt x \lt -\sqrt {4−\epsilon}$ which leads to $2-\sqrt {4+\epsilon} \lt x+2 \lt 2-\sqrt {4−\epsilon}$
Step 5. Compare the result from step 4 with the requirement for the value $\delta$ in step 2; we can set $\delta$ as the smaller value of $|2-\sqrt {4+\epsilon}|$ and $|2-\sqrt {4−\epsilon}|$
Step 6. For $\epsilon\geq4$, choose $\delta$ to be the smaller of $2$ and $|2-\sqrt {4+\epsilon}|$
Step 7. With the setting of the $\delta$ value, we can go backwards through the steps of 4,3,2 to reach a conclusion that for all $x$ when $0\lt |x+2|\lt\delta$, we have $|x^2−4|\lt\epsilon$ which proves the limit statement.
