1. **State the problem:** Find the limit as $x \to -\infty$ of the expression $$\frac{8x + 2}{x - \sqrt{x^2 - 3}}.$$\n\n2. **Recall the formula and rules:** When evaluating limits involving expressions with square roots and polynomials as $x \to \pm \infty$, it is useful to factor out the highest power of $x$ to simplify the expression. Also, remember that for large $|x|$, $\sqrt{x^2 - 3} \approx |x|$ but the sign depends on $x$. Since $x \to -\infty$, $|x| = -x$.\n\n3. **Rewrite the denominator:**\n$$x - \sqrt{x^2 - 3} = x - \sqrt{x^2(1 - \frac{3}{x^2})} = x - |x|\sqrt{1 - \frac{3}{x^2}}.$$\nSince $x \to -\infty$, $|x| = -x$, so\n$$x - \sqrt{x^2 - 3} = x - (-x)\sqrt{1 - \frac{3}{x^2}} = x + x\sqrt{1 - \frac{3}{x^2}} = x(1 + \sqrt{1 - \frac{3}{x^2}}).$$\n\n4. **Rewrite the entire expression:**\n$$\frac{8x + 2}{x - \sqrt{x^2 - 3}} = \frac{8x + 2}{x(1 + \sqrt{1 - \frac{3}{x^2}})} = \frac{8x + 2}{x} \cdot \frac{1}{1 + \sqrt{1 - \frac{3}{x^2}}}.$$\n\n5. **Simplify the numerator fraction:**\n$$\frac{8x + 2}{x} = \frac{\cancel{8x} + 2}{\cancel{x}} = 8 + \frac{2}{x}.$$\n\n6. **Substitute back:**\n$$\left(8 + \frac{2}{x}\right) \cdot \frac{1}{1 + \sqrt{1 - \frac{3}{x^2}}}.$$\n\n7. **Evaluate the limit as $x \to -\infty$:**\n- $\frac{2}{x} \to 0$\n- $\sqrt{1 - \frac{3}{x^2}} \to \sqrt{1 - 0} = 1$\n\nSo the expression tends to\n$$8 \cdot \frac{1}{1 + 1} = 8 \cdot \frac{1}{2} = 4.$$\n\n**Final answer:**\n$$\lim_{x \to -\infty} \frac{8x + 2}{x - \sqrt{x^2 - 3}} = 4.$$