<s>[INST] Evaluating $\lim_{x\to-\infty}\frac{x^3+1}{x^3+\sqrt{4x^6+1}}$ $$\displaystyle \lim_{x\to-\infty}\frac{x^3+1}{x^3+\sqrt{4x^6+1}}$$
I understand that we need to divide by the highest power, but first we want to factor out the $x^6$ from the radical. This gives
$$\displaystyle \lim_{x\to-\infty}\frac{x^3+1}{x^3+\sqrt{x^6}\sqrt{4+\frac{1}{x^6}}}.$$
My confusion is when I plug this limit into a program such as symbolab is that it says this fraction should simplify to $$\displaystyle \lim_{x\to-\infty}\frac{x^3+1}{x^3-x^3\sqrt{4+\frac{1}{x^6}}}$$ because as $x\to-\infty\Rightarrow\sqrt{x^6}=-x^3$. Why is it that we apply this computation here, and not anywhere else in this fraction? I'm mostly confused as to why we need to make this computation in the first place. 
 [/INST] In general, for any real $x$, $\sqrt{x^6}=|x|^3$. As $x\to-\infty$, we may assume that $x<0$ and therefore $\sqrt{x^6}=|x|^3=-x^3$. This simplification allows us to easily divide  the numerator and the denominator by $x^3$, and find that
$$\lim_{x\to-\infty}\frac{x^3+1}{x^3-x^3\sqrt{4+\frac{1}{x^6}}}=\lim_{x\to-\infty}\frac{1+\frac{1}{x^3}}{1-\sqrt{4+\frac{1}{x^6}}}.$$
Can you take it from here?
P.S. On the other hand, as $x\to+\infty$, we may assume that $x>0$, $\sqrt{x^6}=|x|^3=x^3$ and
$$\lim_{x\to+\infty}\frac{x^3+1}{x^3+\sqrt{4x^6+1}}=\lim_{x\to+\infty}\frac{x^3+1}{x^3+x^3\sqrt{4+\frac{1}{x^6}}}=\lim_{x\to+\infty}\frac{1+\frac{1}{x^3}}{1+\sqrt{4+\frac{1}{x^6}}}.$$
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