Identify the highest power of x in the denominator and the effective highest power in the numerator.

The highest power in the denominator is $x^3$.
In the numerator, we have $\sqrt{1 + 4x^6}$. As $x \to \infty$, the 1 becomes negligible compared to $4x^6$. So, the dominant term is $\sqrt{4x^6} = \sqrt{4}\sqrt{x^6} = 2|x^3|$. Since $x \to \infty$, $x$ is positive, so $|x^3| = x^3$. Thus, the effective highest power in the numerator is $x^3$.

For terms inside the square root in the numerator, dividing by $x^3$ is equivalent to dividing by $\sqrt{x^6}$ (since $x \to \infty$, $x^3 = \sqrt{x^6}$).

Numerator:

Denominator:

As $x \to \infty$:

So the limit becomes:

The final answer is $-2$.