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?
 A: Keep in mind that it is a limit at $-\infty$. Therefore, the way you should deal with the numerator is$$\sqrt{4x^2-2x}=-x\frac{\sqrt{4x^2-2x}}{-x}=-x\sqrt{4-\frac2x}.$$
A: Don't forget that $\sqrt{x^2}=|x|$ and when $x<0$ ($x$ is going to negative infinity), $|x|$ is equivalent to $-x$:
$$
\frac{\sqrt{x^2}\sqrt{4-\frac{2}{x}}}{\sqrt[3]{x^3}\sqrt[3]{1+\frac{1}{x^3}}}=
\frac{|x|\sqrt{4-\frac{2}{x}}}{x\sqrt[3]{1+\frac{1}{x^3}}}=
\frac{-x\sqrt{4-\frac{2}{x}}}{x\sqrt[3]{1+\frac{1}{x^3}}}=\\
-\frac{\sqrt{4-\frac{2}{x}}}{\sqrt[3]{1+\frac{1}{x^3}}}\xrightarrow{x\rightarrow-\infty}-\frac{\sqrt{4-0}}{\sqrt[3]{1+0}}=
-\frac{2}{1}=-2.
$$
A: Note that 
$$\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}= \frac{|x|}{x}\stackrel{x<0}{=}-1$$
A: Dividing both the numerator and the denominator by $x$ is always allowed, but the square root creates a trap:
$$\frac{\sqrt{a}}x=\text{sgn }x\sqrt{\frac a{x^2}}$$ because the square root is always a positive number.
A: It is  faster using equivalents. Recall that a polynomial is asymptotically equivalent to its leading term, so
$$\frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}\sim_{-\infty}\frac{\sqrt{4x^2}}{\sqrt[3]{x^3}}=\frac{2|x|}{x}=-2.$$
