1. **State the problem:** Find the limit $$\lim_{x \to -\infty} \left( \frac{1}{\sqrt{1+x^2}} + x \right).$$\n\n2. **Recall important rules:** As $x \to -\infty$, $x^2$ becomes very large, so $\sqrt{1+x^2} \approx \sqrt{x^2} = |x|$. Since $x$ is negative, $|x| = -x$.\n\n3. **Rewrite the expression using this approximation:**\n$$\frac{1}{\sqrt{1+x^2}} + x \approx \frac{1}{|x|} + x = \frac{1}{-x} + x = -\frac{1}{x} + x.$$\n\n4. **Combine terms over a common denominator:**\n$$-\frac{1}{x} + x = -\frac{1}{x} + \frac{x^2}{x} = \frac{-1 + x^2}{x}.$$\n\n5. **Analyze the limit:** As $x \to -\infty$, $x^2$ dominates $-1$, so numerator $\approx x^2$, denominator $= x$. Thus,\n$$\lim_{x \to -\infty} \frac{x^2 - 1}{x} = \lim_{x \to -\infty} \frac{x^2}{x} = \lim_{x \to -\infty} x = -\infty.$$\n\n6. **More precise approach:** To avoid confusion, rationalize the original expression:\n$$\frac{1}{\sqrt{1+x^2}} + x = \frac{1 + x \sqrt{1+x^2}}{\sqrt{1+x^2}}.$$\nBut this is complicated; instead, multiply numerator and denominator by $\sqrt{1+x^2} - x$ to simplify:\n\n7. **Multiply numerator and denominator by $\sqrt{1+x^2} - x$:**\n$$\left( \frac{1}{\sqrt{1+x^2}} + x \right) \cdot \frac{\sqrt{1+x^2} - x}{\sqrt{1+x^2} - x} = \frac{(1 + x \sqrt{1+x^2})(\sqrt{1+x^2} - x)}{(\sqrt{1+x^2})(\sqrt{1+x^2} - x)}.$$\n\n8. **Simplify denominator:**\n$$ (\sqrt{1+x^2})(\sqrt{1+x^2} - x) = (1+x^2) - x \sqrt{1+x^2}.$$\n\n9. **Simplify numerator:**\n$$ (1 + x \sqrt{1+x^2})(\sqrt{1+x^2} - x) = \sqrt{1+x^2} - x + x (1+x^2) - x^2 \sqrt{1+x^2}.$$\n\n10. **Group terms:**\n$$ (\sqrt{1+x^2} - x^2 \sqrt{1+x^2}) + (x (1+x^2) - x) = \sqrt{1+x^2}(1 - x^2) + x (1+x^2 - 1) = \sqrt{1+x^2}(1 - x^2) + x x^2 = \sqrt{1+x^2}(1 - x^2) + x^3.$$\n\n11. **As $x \to -\infty$, $x^2$ dominates 1, so $1 - x^2 \approx -x^2$, and $\sqrt{1+x^2} \approx |x| = -x$ (since $x$ negative):**\n$$\sqrt{1+x^2}(1 - x^2) \approx (-x)(-x^2) = x^3.$$\n\n12. **So numerator $\approx x^3 + x^3 = 2x^3$, denominator $= (1+x^2) - x \sqrt{1+x^2} \approx x^2 - x(-x) = x^2 + x^2 = 2x^2.$**\n\n13. **Therefore, the expression approximates:**\n$$\frac{2x^3}{2x^2} = x.$$\n\n14. **Finally, as $x \to -\infty$, $x \to -\infty$, so the limit is:**\n$$\boxed{-\infty}.$$