# 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.

Thank you.

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 $$a=x^3$$.

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

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

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

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