<s>[INST] Evaluate : $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$ 
Evaluate: $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$

My solution:
\begin{align}
\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}} & = \lim_{x\to -\infty} \frac{\sqrt{4-\frac{2}{x}}}{\sqrt[3]{1+\frac{1}{x^3}}}\\
& = \frac{\sqrt{4-0}}{\sqrt[3]{1+0}} \\
& = 2
\end{align}
Despite the steps I've taken seems plausible to me, the answer is given as $-2$. 
Is dividing both the numerator and the denominator by $x$ allowed here? Where am I making a mistake?
 [/INST] It is  faster using equivalents. Recall that a polynomial is asymptotically equivalent to its leading term, so
$$\frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}\sim_{-\infty}\frac{\sqrt{4x^2}}{\sqrt[3]{x^3}}=\frac{2|x|}{x}=-2.$$
</s>