1. The problem states that $L = \{a \in [-3, \infty) : \lim_{x \to a^-} f(x) \text{ exists}\}$ and $R = \{a \in [-3, \infty) : \lim_{x \to a^+} f(x) \text{ exists}\}$. We need to find $L \cup R$ and $L \cap R$. 2. The set $L$ consists of all points $a$ from $-3$ to $\infty$ where the left-hand limit (limit of $f(x)$ as $x$ approaches $a$ from values less than $a$) exists. 3. The set $R$ consists of all points $a$ from $-3$ to $\infty$ where the right-hand limit (limit of $f(x)$ as $x$ approaches $a$ from values greater than $a$) exists. 4. Typically, for most points $a$ inside an interval, both left-hand and right-hand limits exist unless $a$ is an endpoint or a point of discontinuity where one-sided limits don't exist. 5. Since the domain considered is $[-3, \infty)$, at the left endpoint $a = -3$, there is no left-hand side from the domain because $x \to -3^-$ is outside the domain. 6. Therefore, $L$ cannot contain $-3$ because $\lim_{x \to -3^-} f(x)$ cannot exist as there are no $x < -3$ in the domain. 7. However, the right-hand limit at $-3$ exists if $f(x)$ is defined to the right of $-3$. Thus $-3 \in R$. 8. For all $a > -3$, both $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$ can exist (not given otherwise). So for all $a > -3$, $a \in L$ and $a \in R$. 9. Therefore, $L = (-3, \infty)$ and $R = [-3, \infty)$. 10. The union $L \cup R = (-3, \infty) \cup [-3, \infty) = [-3, \infty)$. 11. The intersection $L \cap R = (-3, \infty) \cap [-3, \infty) = (-3, \infty)$. Final answers: $$L \cup R = [-3, \infty)$$ $$L \cap R = (-3, \infty)$$