1. State the problem: Compute $\lim_{x\to 4}\frac{\sqrt{2x+8}-4}{x-4}$.\n\n2. Observe that direct substitution gives $\frac{0}{0}$, an indeterminate form, so we must manipulate the expression algebraically.\n\n3. Multiply numerator and denominator by the conjugate $\sqrt{2x+8}+4$ to eliminate the square root.\n\n4. Compute the product.\n$$\frac{\sqrt{2x+8}-4}{x-4}\cdot\frac{\sqrt{2x+8}+4}{\sqrt{2x+8}+4}=\frac{2x+8-16}{(x-4)(\sqrt{2x+8}+4)}$$\n\n5. Simplify the numerator: $2x+8-16=2x-8=2(x-4)$.\n\n6. Cancel the common factor $(x-4)$ to get $\frac{2}{\sqrt{2x+8}+4}$.\n\n7. Now take the limit by direct substitution: $\lim_{x\to 4}\frac{2}{\sqrt{2x+8}+4}=\frac{2}{\sqrt{16}+4}=\frac{2}{4+4}=\frac{2}{8}=\frac{1}{4}$.\n\nAnswer: $\frac{1}{4}$.\n