1. **State the problem:** We need to find the limit as $x$ approaches $+\infty$ of the expression $\left(\sqrt{x^2 + 2} - x\right)^5$. 2. **Recall the formula and approach:** When dealing with limits involving expressions like $\sqrt{x^2 + a} - x$ as $x \to +\infty$, it is useful to rationalize the expression to simplify it. 3. **Rationalize the inner expression:** $$\sqrt{x^2 + 2} - x = \frac{(\sqrt{x^2 + 2} - x)(\sqrt{x^2 + 2} + x)}{\sqrt{x^2 + 2} + x} = \frac{x^2 + 2 - x^2}{\sqrt{x^2 + 2} + x} = \frac{2}{\sqrt{x^2 + 2} + x}$$ 4. **Simplify the denominator for large $x$:** As $x \to +\infty$, $\sqrt{x^2 + 2} = x\sqrt{1 + \frac{2}{x^2}} \approx x\left(1 + \frac{1}{2} \cdot \frac{2}{x^2}\right) = x + \frac{1}{x}$. So, $$\sqrt{x^2 + 2} + x \approx x + \frac{1}{x} + x = 2x + \frac{1}{x}$$ 5. **Evaluate the inner expression limit:** $$\frac{2}{2x + \frac{1}{x}} = \frac{2}{2x + \frac{1}{x}} = \frac{2}{2x} \text{ as } x \to +\infty \text{ since } \frac{1}{x} \to 0$$ Thus, $$\sqrt{x^2 + 2} - x \approx \frac{2}{2x} = \frac{1}{x}$$ 6. **Raise to the fifth power:** $$\left(\sqrt{x^2 + 2} - x\right)^5 \approx \left(\frac{1}{x}\right)^5 = \frac{1}{x^5}$$ 7. **Find the limit:** As $x \to +\infty$, $\frac{1}{x^5} \to 0$. **Final answer:** $$\lim_{x \to +\infty} \left(\sqrt{x^2 + 2} - x\right)^5 = 0$$