1. **State the problem:** Find the limit $$\lim_{x \to -\infty} \frac{1}{\sqrt{1 + x^2} + x}$$. 2. **Recall the formula and rules:** When dealing with limits involving infinity and square roots, a common technique is to rationalize the expression or factor out the dominant term to simplify. 3. **Rewrite the expression:** To simplify, multiply numerator and denominator by the conjugate of the denominator: $$\frac{1}{\sqrt{1 + x^2} + x} \times \frac{\sqrt{1 + x^2} - x}{\sqrt{1 + x^2} - x} = \frac{\sqrt{1 + x^2} - x}{(\sqrt{1 + x^2} + x)(\sqrt{1 + x^2} - x)}$$ 4. **Simplify the denominator using difference of squares:** $$= \frac{\sqrt{1 + x^2} - x}{(1 + x^2) - x^2} = \frac{\sqrt{1 + x^2} - x}{1} = \sqrt{1 + x^2} - x$$ 5. **Evaluate the limit:** As $x \to -\infty$, $x$ is very large negative, so $|x| = -x$. Rewrite: $$\sqrt{1 + x^2} - x = |x| \sqrt{1 + \frac{1}{x^2}} - x = -x \sqrt{1 + \frac{1}{x^2}} - x$$ 6. **Factor out $-x$:** $$= -x \left( \sqrt{1 + \frac{1}{x^2}} + 1 \right)$$ 7. **As $x \to -\infty$, $\frac{1}{x^2} \to 0$, so:** $$\sqrt{1 + \frac{1}{x^2}} \to 1$$ 8. **Therefore:** $$\lim_{x \to -\infty} \sqrt{1 + x^2} - x = \lim_{x \to -\infty} -x (1 + 1) = \lim_{x \to -\infty} -2x$$ 9. **Since $x \to -\infty$, $-2x \to +\infty$.** **Final answer:** $$\lim_{x \to -\infty} \frac{1}{\sqrt{1 + x^2} + x} = +\infty$$