1. **State the problem:** Find the limit $$\lim_{x \to +\infty} \sqrt[3]{x^2 + x + 1}$$. 2. **Recall the cube root limit rule:** For large $x$, the term with the highest power dominates inside the root. Here, $x^2$ dominates $x$ and $1$. 3. **Rewrite the expression inside the cube root:** $$\sqrt[3]{x^2 + x + 1} = \sqrt[3]{x^2 \left(1 + \frac{1}{x} + \frac{1}{x^2}\right)}$$ 4. **Use the property of roots:** $$= \sqrt[3]{x^2} \cdot \sqrt[3]{1 + \frac{1}{x} + \frac{1}{x^2}}$$ 5. **Simplify the cube root of $x^2$:** $$\sqrt[3]{x^2} = x^{\frac{2}{3}}$$ 6. **Evaluate the limit of the second factor as $x \to +\infty$:** Since $\frac{1}{x} \to 0$ and $\frac{1}{x^2} \to 0$, $$\lim_{x \to +\infty} \sqrt[3]{1 + \frac{1}{x} + \frac{1}{x^2}} = \sqrt[3]{1} = 1$$ 7. **Combine the results:** $$\lim_{x \to +\infty} \sqrt[3]{x^2 + x + 1} = \lim_{x \to +\infty} x^{\frac{2}{3}} \cdot 1 = +\infty$$ **Final answer:** $$\boxed{+\infty}$$