1. **State the problem:** Evaluate the limit \( \lim_{x \to 4} \left( \frac{\sqrt{2x+8}}{4} - 2 \right) \div \left( \sqrt{2x+8} - 4 \right) \).\n\n2. **Simplify the expression:**\nRewrite limit as \( \lim_{x \to 4} \frac{ \frac{\sqrt{2x+8}}{4} - 2 }{ \sqrt{2x+8} - 4 } \).\n\n3. **Find common denominator in numerator:**\n\( \frac{\sqrt{2x+8}}{4} - 2 = \frac{\sqrt{2x+8} - 8}{4} \).\n\n4. **Substitute numerator back:**\nThe limit becomes \( \lim_{x \to 4} \frac{ \frac{\sqrt{2x+8} - 8}{4} }{ \sqrt{2x+8} - 4 } = \lim_{x \to 4} \frac{\sqrt{2x+8} - 8}{4(\sqrt{2x+8} - 4)} \).\n\n5. **Factor numerator:**\nNote that \( \sqrt{2x+8} - 8 = (\sqrt{2x+8} - 4)(\sqrt{2x+8} + 4) \) because \( (\sqrt{2x+8} -4)(\sqrt{2x+8} +4) = (\sqrt{2x+8})^2 - 16 = 2x+8 - 16 = 2x-8 \) but this doesn't exactly equal \( \sqrt{2x+8} - 8 \), so let's check again.\n\nActually, \( \sqrt{2x+8} - 8 \) cannot be factored like that; false step. Instead, rewrite to handle the denominator.\n\n6. **Use substitution:** Let \( t = \sqrt{2x+8} \). When \( x \to 4, t = \sqrt{2 \cdot 4 + 8} = \sqrt{16} = 4 \).\nThen the limit becomes \( \lim_{t \to 4} \frac{t - 8}{4 (t - 4)} \).\n\n7. **Simplify the numerator:** \( t -8 = (t -4) - 4 \). So \( \frac{t -8}{4 (t -4)} = \frac{(t-4) -4}{4(t-4)} = \frac{t-4}{4(t-4)} - \frac{4}{4(t-4)} = \frac{1}{4} - \frac{1}{t-4} \) which diverges as \( t \to 4 \). So this is not valid. So we must reconsider and carefully rationalize the denominator or numerator.\n\n8. **Alternative approach: multiply numerator and denominator by conjugate of denominator:**\nMultiply by \( \frac{\sqrt{2x+8} + 4}{\sqrt{2x+8} + 4} \).\n\nExpression becomes:\n\( \frac{\frac{\sqrt{2x+8}}{4} - 2}{\sqrt{2x+8} -4} \cdot \frac{\sqrt{2x+8} + 4}{\sqrt{2x+8} + 4} = \frac{\left( \frac{\sqrt{2x+8}}{4} - 2 \right) (\sqrt{2x+8} + 4)}{(\sqrt{2x+8} -4)(\sqrt{2x+8} +4)} \).\n\n9. **Simplify denominator:** \( (\sqrt{2x+8} -4)(\sqrt{2x+8} +4) = 2x+8 - 16 = 2x -8 \).\n\n10. **Rewrite entire expression:**\n\( \frac{\left( \frac{\sqrt{2x+8}}{4} - 2 \right) (\sqrt{2x+8} + 4)}{2x -8} = \frac{\left( \frac{\sqrt{2x+8}}{4} - 2 \right) (\sqrt{2x+8} + 4)}{2(x -4)} \).\n\n11. **Simplify numerator factor inside first parenthesis:**\n\( \frac{\sqrt{2x+8}}{4} - 2 = \frac{\sqrt{2x+8} - 8}{4} \).\n\n12. **Substitute back:**\n\( \frac{ (\sqrt{2x+8} - 8)(\sqrt{2x+8} + 4) }{4 \cdot 2 (x-4)} = \frac{ (\sqrt{2x+8} - 8)(\sqrt{2x+8} +4) }{8 (x-4)} \).\n\n13. **Multiply terms in numerator:**\nUse \( (a - b)(a + c) = a^2 + ac - ab - bc \)\n\( (\sqrt{2x+8} -8)(\sqrt{2x+8} +4) = (\sqrt{2x+8})^2 + 4 \sqrt{2x+8} - 8 \sqrt{2x+8} - 32 = 2x+8 - 4 \sqrt{2x+8} - 32 \).\n\n14. **Simplify numerator:**\n\( 2x + 8 - 4 \sqrt{2x+8} - 32 = 2x -24 - 4 \sqrt{2x+8} \).\n\n15. **Rewrite expression:**\n\( \frac{2x - 24 - 4 \sqrt{2x+8}}{8(x - 4)} \).\n\n16. **Factor numerator:**\nNotice \( 2x - 24 = 2(x - 12) \), so numerator is \( 2(x - 12) - 4 \sqrt{2x+8} \). Cannot factor easily; instead, substitute \( x = 4 + h \) and take limit as \( h \to 0 \).\n\n17. **Substitute \( x = 4 + h \):**\nExpression becomes \( \frac{2(4+h) - 24 - 4 \sqrt{2(4+h)+8}}{8((4+h)-4)} = \frac{8 + 2h - 24 - 4 \sqrt{8 + 2h + 8}}{8 h} = \frac{-16 + 2h - 4 \sqrt{16 + 2h}}{8 h} \).\n\n18. **Rewrite numerator:**\n\( -16 + 2h - 4 \sqrt{16 + 2h} = -4 \left(4 - \frac{h}{2} + \sqrt{16 + 2h} \right) \), but simpler to approximate or use expansion.\n\n19. **Use series expansion for \( \sqrt{16 + 2h} = 4 + \frac{2h}{2 \times 4} + ... = 4 + \frac{h}{4} + ... \).\n\n20. **Plug back approximation:**\n\( -16 + 2h - 4\left(4 + \frac{h}{4} \right) = -16 + 2h - 16 - h = -32 + h \).\n\n21. **Expression is:**\n\( \frac{-32 + h}{8 h} = \frac{-32}{8 h} + \frac{h}{8 h} = -\frac{4}{h} + \frac{1}{8} \) which diverges as \( h \to 0 \). So the limit does not exist in finite form this way.\n\n22. **Reconsider the original problem interpretation.** Possibly parentheses or formatting is unclear. Re-express original problem to clarify.\n\nGiven problem: \( \lim_{x \to 4} \left( \frac{\sqrt{2x+8}}{4} - 2 \right) \div \left( \sqrt{2x+8} - 4 \right) \).\n\nAt \( x=4 \), numerator \( \frac{4}{4} - 2 = 1-2 = -1 \), denominator \( 4 -4 = 0 \) causes division by zero.\n\nHence the limit tends to \( \pm \infty \) or does not exist.\n\n23. **Check limits from left and right:**\nFor \( x \to 4^+ \), \( \sqrt{2x+8} > 4 \), denominator positive, numerator negative (around -1), quotient tends to \(-\infty\).\nFor \( x \to 4^- \), denominator negative, numerator still about -1, quotient tends to \+\infty\).\n\n24. **Conclusion:** The limit does not exist because left and right limits differ.\n\n**Final answer:** \n$$\lim_{x \to 4} \frac{\frac{\sqrt{2x+8}}{4} - 2}{\sqrt{2x+8} - 4} \text{ does not exist}.$$