1. **State the problem:** Find the limit $$\lim_{x \to -\infty} \frac{\sqrt{x^4 + 4x + 1}}{\sqrt{x^2 + 1}}.$$\n\n2. **Recall the formula and rules:** When dealing with limits at infinity involving square roots and polynomials, factor out the highest power of $x$ inside the root to simplify.\n\n3. **Simplify numerator:** Inside the numerator's square root, the highest power is $x^4$. Factor it out:\n$$\sqrt{x^4 + 4x + 1} = \sqrt{x^4\left(1 + \frac{4x}{x^4} + \frac{1}{x^4}\right)} = \sqrt{x^4} \sqrt{1 + \frac{4}{x^3} + \frac{1}{x^4}} = |x^2| \sqrt{1 + \frac{4}{x^3} + \frac{1}{x^4}}.$$\nSince $x \to -\infty$, $x^2$ is positive and $|x^2| = x^2$.\n\n4. **Simplify denominator:** Inside the denominator's square root, the highest power is $x^2$. Factor it out:\n$$\sqrt{x^2 + 1} = \sqrt{x^2\left(1 + \frac{1}{x^2}\right)} = |x| \sqrt{1 + \frac{1}{x^2}}.$$\nSince $x \to -\infty$, $|x| = -x$.\n\n5. **Rewrite the limit:**\n$$\lim_{x \to -\infty} \frac{\sqrt{x^4 + 4x + 1}}{\sqrt{x^2 + 1}} = \lim_{x \to -\infty} \frac{x^2 \sqrt{1 + \frac{4}{x^3} + \frac{1}{x^4}}}{|x| \sqrt{1 + \frac{1}{x^2}}} = \lim_{x \to -\infty} \frac{x^2}{|x|} \cdot \frac{\sqrt{1 + \frac{4}{x^3} + \frac{1}{x^4}}}{\sqrt{1 + \frac{1}{x^2}}}.$$\n\n6. **Simplify the fraction of powers:**\n$$\frac{x^2}{|x|} = \frac{x^2}{-x} = -x.$$\n\n7. **Evaluate the limit of the square root expressions:** As $x \to -\infty$, terms with $\frac{1}{x^n}$ go to zero, so:\n$$\sqrt{1 + \frac{4}{x^3} + \frac{1}{x^4}} \to \sqrt{1} = 1,$$\n$$\sqrt{1 + \frac{1}{x^2}} \to \sqrt{1} = 1.$$\n\n8. **Combine all parts:**\n$$\lim_{x \to -\infty} -x \cdot \frac{1}{1} = \lim_{x \to -\infty} -x = +\infty.$$\n\n**Final answer:** $$\boxed{+\infty}.$$