<s>[INST] limit of $\sqrt{x^6}$ as $x$ approaches $-\infty$ I need to solve this limit:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}}$$
The answer is $-3$, but I got 3 instead. This is my process:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}} =
\lim_{x \to - \infty}{\frac {\sqrt{x^6(9-\frac {5}{x^2})}}{x^3(1-\frac {2}{x}+\frac{1}{x^3})}} =
\lim_{x \to - \infty}{\frac {\sqrt{x^6}\sqrt{(9-\frac {5}{x^2})}}{x^3(1-\frac {2}{x}+\frac{1}{x^3})}} =
\lim_{x \to - \infty}{\frac {\require{cancel} \cancel{x^3} \sqrt{(9-\frac {5}{x^2})}}{\require{cancel} \cancel{x^3}(1-\frac {2}{x}+\frac{1}{x^3})}} =
\frac {3}{1} = +3$$
I've been told that in the third step the $\sqrt{x^6}$ should be equal $\textbf{-}\sqrt{x^3}$, but I didn't understand why.
I'll be glad to get your help!
Thank you.
 [/INST] You are forgetting that $\sqrt{x^6}=|x^3|$ and $\frac{|x^3|}{x^3}=-1$ when $x$ is negative
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