1. **State the problem:** We need to find the limit $$\lim_{x \to -\infty} \frac{2x - 4x^3}{6x^2 + e^{-x}}.$$\n\n2. **Identify dominant terms:** As $x \to -\infty$, the term $-4x^3$ dominates the numerator because it grows faster than $2x$. In the denominator, $6x^2$ and $e^{-x}$ compete, but since $e^{-x} = e^{|x|}$ grows exponentially as $x \to -\infty$, it dominates $6x^2$.\n\n3. **Rewrite the limit focusing on dominant terms:**\n$$\lim_{x \to -\infty} \frac{-4x^3}{e^{-x}}.$$\n\n4. **Analyze the behavior:**\n- Numerator: $-4x^3$ tends to $-\infty$ as $x \to -\infty$.\n- Denominator: $e^{-x} = e^{|x|}$ tends to $+\infty$ much faster than any polynomial.\n\n5. **Evaluate the limit:** Since the denominator grows exponentially and numerator grows polynomially, the fraction tends to zero. The sign depends on numerator and denominator signs, but denominator is positive and numerator negative for large negative $x$, so the fraction tends to $0$ from the negative side.\n\n6. **Final answer:**\n$$\lim_{x \to -\infty} \frac{2x - 4x^3}{6x^2 + e^{-x}} = 0.$$