1. **State the problem:** Find the limit $$\lim_{x \to -\infty} 2\sqrt{x^2+3} - x - 3.$$\n\n2. **Recall the formula and rules:** When $x \to -\infty$, $x^2$ dominates inside the square root. We use the technique of factoring $x^2$ inside the root to simplify:\n$$\sqrt{x^2+3} = \sqrt{x^2\left(1 + \frac{3}{x^2}\right)} = |x|\sqrt{1 + \frac{3}{x^2}}.$$\nSince $x \to -\infty$, $|x| = -x$.\n\n3. **Rewrite the expression:**\n$$2\sqrt{x^2+3} - x - 3 = 2(-x)\sqrt{1 + \frac{3}{x^2}} - x - 3 = -2x\sqrt{1 + \frac{3}{x^2}} - x - 3.$$\n\n4. **Use binomial expansion for the square root for large $|x|$:**\n$$\sqrt{1 + \frac{3}{x^2}} \approx 1 + \frac{1}{2} \cdot \frac{3}{x^2} = 1 + \frac{3}{2x^2}.$$\n\n5. **Substitute approximation:**\n$$-2x\left(1 + \frac{3}{2x^2}\right) - x - 3 = -2x - \frac{3}{x} - x - 3 = (-2x - x) - \frac{3}{x} - 3 = -3x - \frac{3}{x} - 3.$$\n\n6. **Evaluate the limit as $x \to -\infty$:**\n- The term $-3x$ tends to $+\infty$ because $x$ is large negative and multiplied by $-3$.\n- The term $-\frac{3}{x}$ tends to 0.\n- The constant $-3$ remains $-3$.\n\nTherefore, the whole expression tends to $+\infty$.\n\n**Final answer:**\n$$\lim_{x \to -\infty} 2\sqrt{x^2+3} - x - 3 = +\infty.$$