1. **State the problem:** We want to find the limit as $x$ approaches positive infinity of the expression $$\left( \frac{\sqrt[x]{2} + \sqrt[x]{4} + \sqrt[x]{8}}{3} \right)^x.$$\n\n2. **Recall the formula and rules:** For limits of the form $$\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^x = e^a,$$ we often try to rewrite the expression inside the parentheses to look like $1 + \frac{a}{x}$.\n\n3. **Analyze the terms inside the parentheses:** Note that $$\sqrt[x]{2} = 2^{1/x}, \quad \sqrt[x]{4} = 4^{1/x} = (2^2)^{1/x} = 2^{2/x}, \quad \sqrt[x]{8} = 8^{1/x} = (2^3)^{1/x} = 2^{3/x}.$$\n\n4. **Rewrite the sum:** $$\sqrt[x]{2} + \sqrt[x]{4} + \sqrt[x]{8} = 2^{1/x} + 2^{2/x} + 2^{3/x}.$$\n\n5. **Use the exponential approximation for large $x$:** For large $x$, $$2^{k/x} = e^{(k/x) \ln 2} \approx 1 + \frac{k \ln 2}{x}$$ for $k=1,2,3$.\n\n6. **Sum the approximations:** $$2^{1/x} + 2^{2/x} + 2^{3/x} \approx \left(1 + \frac{\ln 2}{x}\right) + \left(1 + \frac{2 \ln 2}{x}\right) + \left(1 + \frac{3 \ln 2}{x}\right) = 3 + \frac{6 \ln 2}{x}.$$\n\n7. **Divide by 3:** $$\frac{\sqrt[x]{2} + \sqrt[x]{4} + \sqrt[x]{8}}{3} \approx \frac{3 + \frac{6 \ln 2}{x}}{3} = 1 + \frac{2 \ln 2}{x}.$$\n\n8. **Rewrite the original limit:** $$\lim_{x \to \infty} \left(1 + \frac{2 \ln 2}{x}\right)^x.$$\n\n9. **Apply the exponential limit formula:** $$\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^x = e^a,$$ so here $$a = 2 \ln 2,$$ thus the limit is $$e^{2 \ln 2}.$$\n\n10. **Simplify the exponent:** $$e^{2 \ln 2} = (e^{\ln 2})^2 = 2^2 = 4.$$\n\n**Final answer:** $$\boxed{4}.$$