1. **State the problem:** Find the limit as $n \to \infty$ of $$\frac{\sqrt{n^{2}+5} + \sqrt{n}}{\sqrt[3]{8n^{3} + 2n - n}}.$$\n\n2. **Simplify the numerator:** For large $n$, $\sqrt{n^{2}+5} = \sqrt{n^{2}(1 + \frac{5}{n^{2}})} = n \sqrt{1 + \frac{5}{n^{2}}}.$ As $n \to \infty$, $\sqrt{1 + \frac{5}{n^{2}}} \to 1$, so $\sqrt{n^{2} + 5} \sim n.$\n\n3. The second term in numerator is $\sqrt{n} = n^{1/2}.$\n\n4. Therefore, the numerator behaves like $n + n^{1/2}$ as $n \to \infty.$ Since $n$ grows faster than $n^{1/2}$, the dominant term in the numerator is $n.$\n\n5. **Simplify the denominator:**\n$$\sqrt[3]{8n^{3} + 2n - n} = \sqrt[3]{8n^{3} + n} = \sqrt[3]{n^{3}(8 + \frac{1}{n^{2}})} = n \sqrt[3]{8 + \frac{1}{n^{2}}}.$$\n\nAs $n \to \infty$, $\sqrt[3]{8 + \frac{1}{n^{2}}} \to \sqrt[3]{8} = 2$.\n\n6. So the denominator behaves like $n \times 2 = 2n.$\n\n7. **Combine numerator and denominator:**\n$$\frac{n + n^{1/2}}{2n} = \frac{n}{2n} + \frac{n^{1/2}}{2n} = \frac{1}{2} + \frac{1}{2} n^{-1/2}.$$\n\n8. **Evaluate the limit:**\nAs $n \to \infty$, $n^{-1/2} \to 0$, so the limit is $$\lim_{n \to \infty} \frac{\sqrt{n^{2}+5}+ \sqrt{n}}{\sqrt[3]{8n^{3}+2n-n}} = \frac{1}{2} + 0 = \frac{1}{2}.$$\n\n**Final answer:** $$\boxed{\frac{1}{2}}.$$