Answer
See explanations.
Work Step by Step
Step 1. We can identify the quantities as: $f(x)=\frac{x^2-9}{x+3},c=-3,L=-6$. Since when $x\to-3, x\ne-3$, we can simply the function as $f(x)=\frac{x^2-9}{x+3}=\frac{(x-3)(x+3)}{x+3}=x-3$
Step 2. For a given small value $\epsilon\gt0$, we need to find a value $\delta\gt0$ so that for all $x$ when $0\lt |x+3|\lt\delta$, we have $|f(x)+6|\lt\epsilon$ or $|x+3|\lt\epsilon$
Step 3. The last inequality gives $-\epsilon\lt x+3\lt\epsilon$ or $-3-\epsilon\lt x \lt\epsilon-3$
Step 4. From the requirement for the value $\delta$ in step 2, we have $-\delta\lt x+3\lt\delta$ or $-3-\delta\lt x\lt\delta-3$
Step 5. Compare the results from steps 4 and 5, we can simply set $\delta=\epsilon$
Step 6. With the setting of the $\delta$ value, we can go backwards through the above steps to reach a conclusion that for all $x$ when $0\lt |x+3|\lt\delta$, we have $|f(x)+6|\lt\epsilon$ which proves the limit statement.
