1. **State the problem:** Find the limit $$\lim_{x \to -\infty} \frac{1}{\overline{\sqrt{1+x^2}} + x}$$ where the bar indicates the conjugate expression. 2. **Recall the conjugate multiplication technique:** To simplify expressions with square roots in the denominator, multiply numerator and denominator by the conjugate of the denominator. 3. **Multiply numerator and denominator by the conjugate:** $$\frac{1}{\overline{\sqrt{1+x^2}} + x} \times \frac{\overline{\sqrt{1+x^2}} - x}{\overline{\sqrt{1+x^2}} - x} = \frac{\overline{\sqrt{1+x^2}} - x}{(\overline{\sqrt{1+x^2}} + x)(\overline{\sqrt{1+x^2}} - x)}$$ 4. **Simplify the denominator using difference of squares:** $$= \frac{\overline{\sqrt{1+x^2}} - x}{(\overline{\sqrt{1+x^2}})^2 - x^2} = \frac{\overline{\sqrt{1+x^2}} - x}{1 + x^2 - x^2} = \frac{\overline{\sqrt{1+x^2}} - x}{1} = \overline{\sqrt{1+x^2}} - x$$ 5. **Evaluate the limit:** As $x \to -\infty$, $\sqrt{1+x^2} \approx |x| = -x$ (since $x$ is negative), so $$\overline{\sqrt{1+x^2}} - x = \sqrt{1+x^2} - x \approx -x - x = -2x$$ 6. **Since $x \to -\infty$, $-2x \to +\infty$.** **Final answer:** $$\lim_{x \to -\infty} \frac{1}{\overline{\sqrt{1+x^2}} + x} = +\infty$$