1. **State the problem:** Find the possible values of $k$ such that the limit $\lim_{x \to k} f(x)$ exists for the piecewise function: $$ f(x) = \begin{cases} 5 - p^2, & x \leq 2 \\ \frac{x - 2}{\sqrt{2x} - 2}, & 2 < x \leq k \\ x - 5, & x > k \end{cases} $$ 2. **Understand the problem:** For the limit $\lim_{x \to k} f(x)$ to exist, the left-hand limit and right-hand limit at $x = k$ must be equal. 3. **Find the left-hand limit at $x = k$:** Since for $x \leq k$, $f(x) = \frac{x - 2}{\sqrt{2x} - 2}$ (because $k \geq 2$), the left-hand limit is: $$ \lim_{x \to k^-} f(x) = \lim_{x \to k^-} \frac{x - 2}{\sqrt{2x} - 2} $$ 4. **Simplify the expression for the left-hand limit:** Multiply numerator and denominator by the conjugate $\sqrt{2x} + 2$: $$ \frac{x - 2}{\sqrt{2x} - 2} \times \frac{\sqrt{2x} + 2}{\sqrt{2x} + 2} = \frac{(x - 2)(\sqrt{2x} + 2)}{(\sqrt{2x})^2 - 2^2} = \frac{(x - 2)(\sqrt{2x} + 2)}{2x - 4} $$ Since $2x - 4 = 2(x - 2)$, this simplifies to: $$ \frac{(x - 2)(\sqrt{2x} + 2)}{2(x - 2)} = \frac{\sqrt{2x} + 2}{2}, \quad x \neq 2 $$ 5. **Evaluate the left-hand limit at $x = k$:** $$ \lim_{x \to k^-} f(x) = \frac{\sqrt{2k} + 2}{2} $$ 6. **Find the right-hand limit at $x = k$:** For $x > k$, $f(x) = x - 5$, so: $$ \lim_{x \to k^+} f(x) = k - 5 $$ 7. **Set the left and right limits equal for the limit to exist:** $$ \frac{\sqrt{2k} + 2}{2} = k - 5 $$ 8. **Solve for $k$:** Multiply both sides by 2: $$ \sqrt{2k} + 2 = 2k - 10 $$ Rearranged: $$ \sqrt{2k} = 2k - 12 $$ 9. **Square both sides:** $$ 2k = (2k - 12)^2 = 4k^2 - 48k + 144 $$ 10. **Bring all terms to one side:** $$ 0 = 4k^2 - 48k + 144 - 2k = 4k^2 - 50k + 144 $$ 11. **Divide entire equation by 2 for simplicity:** $$ 0 = 2k^2 - 25k + 72 $$ 12. **Use quadratic formula:** $$ k = \frac{25 \pm \sqrt{25^2 - 4 \times 2 \times 72}}{2 \times 2} = \frac{25 \pm \sqrt{625 - 576}}{4} = \frac{25 \pm \sqrt{49}}{4} $$ 13. **Calculate roots:** $$ k = \frac{25 \pm 7}{4} $$ So, $$ k_1 = \frac{25 + 7}{4} = 8, \quad k_2 = \frac{25 - 7}{4} = 4.5 $$ 14. **Check domain restrictions:** Since $k$ must be greater than or equal to 2 (to include the middle piece), both $k=4.5$ and $k=8$ are valid. **Final answer:** $$ \boxed{k = 4.5 \text{ or } k = 8} $$