Question: $$\lim\_{x\to-\infty}\frac{\sqrt{x^6+729}}{4x^3+\sqrt{2x^6+1}}=\lim\_{x\to-\infty}\frac{-\sqrt{1+\frac{729}{x^6}}}{4-\sqrt{2+\frac{1}{x^6}}}=\frac{-1}{4-\sqrt{2}}$$
$$\lim\_{x\to\infty}\frac{\sqrt{x^6+729}}{4x^3+\sqrt{2x^6+1}}=\lim\_{x\to-\infty}\frac{\sqrt{1+\frac{729}{x^6}}}{4+\sqrt{2+\frac{1}{x^6}}}=\frac{1}{4+\sqrt{2}}$$
Is this correct? Can someone please explain to me the distribution of the negatives in regards to the radicals when dealing with limits to $-\infty?$ Does it depend on the powers whether or not a negative stays outside of the radicals? Also, why does a negative not get applied to the $4$? Any help would be greatly appreciated!

Answer: 
Your result is correct. 

About the sign of the radical for $ x \to -\infty$, remember that: $\sqrt{X^2}=|X|$, by definition, so, if $X<0$ we have $\sqrt{X^2}=-X$
@Nick, here is how I did it:$$\lim\_{t\to\infty}\frac{\sqrt{t^6+729}}{-4t^3+\sqrt{2t^6+1}}=\lim\_{t\to\infty}\frac{\sqrt{1+\frac{729}{t^6}}}{-4+\sqrt{2+\frac{1}{t^6}}}=\frac{1}{-4+\sqrt{2}}$$