<s>[INST] Limit at negative infinity Evaluate $$\lim_{x\to -\infty}\frac{\sqrt[3]{x^6+8}}{4x^2+\sqrt{3x^4+1}}$$
I think the strategy is to divide the numerator and denominator by x^2. Help please. The textbook answer is $$\frac{1}{4+\sqrt{3}}$$
 [/INST] Instead of dividing the numerator and denominator by $x^2$, it may be easier to factor out the appropriate factors from the radical expressions:
\begin{align}
\frac{\sqrt[3]{x^6+8}}{4x^2+\sqrt{3x^4+1}} & = \frac{\sqrt[3]{x^6\big(1+\frac{8}{x^6}\big)}}{4x^2+\sqrt{x^4\big(3+\frac{1}{x^4}\big)}} \\ & = \frac{\sqrt[3]{x^6}\sqrt[3]{1+\frac{8}{x^6}}}{4x^2+\sqrt{x^4}\sqrt{3+\frac{1}{x^4}}} \\ & = \frac{x^2\sqrt[3]{1+\frac{8}{x^6}}}{4x^2+x^2\sqrt{3+\frac{1}{x^4}}}
\end{align}
</s>