1. **State the problem:** We need to find the value of $k$ such that the limit of the function $$f(x) = \begin{cases} x^2 - k, & x < -3 \\ x^2 + 3, & x \geq -3 \end{cases}$$ exists at $x = -3$. 2. **Recall the limit existence condition:** For the limit to exist at $x = -3$, the left-hand limit and right-hand limit must be equal: $$\lim_{x \to -3^-} f(x) = \lim_{x \to -3^+} f(x)$$ 3. **Calculate the left-hand limit:** $$\lim_{x \to -3^-} f(x) = \lim_{x \to -3^-} (x^2 - k) = (-3)^2 - k = 9 - k$$ 4. **Calculate the right-hand limit:** $$\lim_{x \to -3^+} f(x) = \lim_{x \to -3^+} (x^2 + 3) = (-3)^2 + 3 = 9 + 3 = 12$$ 5. **Set the limits equal to find $k$:** $$9 - k = 12$$ 6. **Solve for $k$:** $$k = 9 - 12 = -3$$ **Final answer:** The value of $k$ for which the limit exists at $x = -3$ is $\boxed{-3}$.