1. **State the problem:** We want to find the limit as $n \to \infty$ of the expression $$\frac{\sqrt{n^2+1}+n}{\sqrt[4]{n^3+n}-\sqrt{n}}.$$\n\n2. **Recall important rules:** When $n$ becomes very large, dominant terms in expressions determine the behavior. We will factor out the highest powers of $n$ in numerator and denominator to simplify.\n\n3. **Simplify numerator:**\n$$\sqrt{n^2+1}+n = n\sqrt{1+\frac{1}{n^2}} + n = n\left(\sqrt{1+\frac{1}{n^2}} + 1\right).$$\nAs $n \to \infty$, $\sqrt{1+\frac{1}{n^2}} \to 1$, so numerator behaves like $n(1+1) = 2n$.\n\n4. **Simplify denominator:**\nFirst, write each term factoring out powers of $n$:\n$$\sqrt[4]{n^3+n} = \sqrt[4]{n^3\left(1 + \frac{1}{n^2}\right)} = n^{3/4} \sqrt[4]{1 + \frac{1}{n^2}} \approx n^{3/4} \cdot 1 = n^{3/4}.$$\nSimilarly,\n$$\sqrt{n} = n^{1/2}.$$\nSo denominator is approximately $$n^{3/4} - n^{1/2} = n^{1/2}(n^{1/4} - 1).$$\n\n5. **Rewrite the whole expression:**\n$$\frac{\sqrt{n^2+1}+n}{\sqrt[4]{n^3+n}-\sqrt{n}} \approx \frac{2n}{n^{1/2}(n^{1/4} - 1)} = \frac{2n}{n^{1/2} \cdot n^{1/4} - n^{1/2}} = \frac{2n}{n^{3/4} - n^{1/2}}.$$\n\n6. **Divide numerator and denominator by $n^{3/4}$ to analyze limit:**\n$$\frac{2n}{n^{3/4} - n^{1/2}} = \frac{2n / n^{3/4}}{1 - n^{1/2} / n^{3/4}} = \frac{2n^{1/4}}{1 - n^{-1/4}}.$$\nAs $n \to \infty$, $n^{-1/4} \to 0$, so denominator $\to 1$. Numerator $\to 2 \cdot \infty = \infty$.\n\n7. **Conclusion:** The limit diverges to infinity.\n\n**Final answer:** $$\lim_{n \to \infty} \frac{\sqrt{n^2+1}+n}{\sqrt[4]{n^3+n}-\sqrt{n}} = \infty.$$