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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.

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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.

Because $\sqrt{a^2}=a$ is only true if $a \ge 0$; for $a \le 0$, you have $\sqrt{a^2}=-a$.

You can summarize this as follows (and remember by heart!), for all $a$ you have: $$\boxed{\sqrt{a^2}=|a|}$$ Apply this to $a=x^3$.

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You are forgetting that $\sqrt{x^6}=|x^3|$ and $\frac{|x^3|}{x^3}=-1$ when $x$ is negative

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Observe that if $y < 0$ then $$\sqrt{y^2} = |y| = -y.$$ Take $y=-3$ for example. In your case, $y = x^3$.

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To check or avoid confusion with sign let $y=-x \to \infty$ then

$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}}=\lim_{y \to \infty}{\frac {\sqrt{9y^6+5y}}{-y^3-2y^2+1}}=-3$$

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