Evaluate $ \lim_{x \rightarrow - \infty} \frac{\sqrt{9x^6-x}}{x^3+1} $ I have started learning limits in calculus and I came across this question: 
Evaluate $ \displaystyle \lim_{x \rightarrow \infty} \dfrac{\sqrt{9x^6-x}}{x^3+1} $ . 
I rewrite the above as $ \displaystyle \lim_{x \rightarrow \infty} \dfrac{\sqrt{9-\dfrac{1}{x^5}}}{1+\dfrac{1}{x^3}} = \lim_{x \rightarrow \infty} \dfrac{\sqrt{9}}{1} = \boxed{3}  $
But now I am asked to compute $ \displaystyle \lim_{x \rightarrow -\infty} \dfrac{\sqrt{9x^6-x}}{x^3+1} $
How to solve for minus infinity? Where to put minus sign , I am getting confused , please help. Thanks.
 A: Dividing by $|x^3|$ we get
$$\frac{\sqrt{9x^6-x}}{x^3+1}=\frac{\frac{\sqrt{9x^6-x}}{|x^3|}}{\frac{x^3+1}{|x^3|}}$$
As $x\to-\infty$ we have, for $x<0$, 
\begin{align}
\frac{\frac{\sqrt{9x^6-x}}{|x^3|}}{\frac{x^3+1}{|x^3|}}&=\frac{\sqrt{\frac{9x^6-x}{x^6}}}{\frac{x^3+1}{-x^3}}=\frac{\sqrt{9-\frac{1}{x^5}}}{-1-\frac{1}{x^3}}\to\frac{\sqrt{9+0}}{-1+0}=\color{blue}{-3}
\end{align}
A: Notice, one can easily change limit as $x\to +\infty$ as follows $$\lim_{x\to -\infty}\frac{\sqrt{9x^6-x}}{x^3+1}$$
$$=\lim_{x\to +\infty}\frac{\sqrt{9(-x)^6-(-x)}}{(-x)^3+1}$$
$$=\lim_{x\to +\infty}\frac{\sqrt{9x^6+x}}{1-x^3}$$
$$=\lim_{x\to +\infty}\frac{|3x^3|\sqrt{1+\frac{1}{9x^5}}}{x^3\left(\frac{1}{x^3}-1\right)}$$
$$=\lim_{x\to +\infty}\frac{3x^3\sqrt{1+\frac{1}{9x^5}}}{x^3\left(\frac{1}{x^3}-1\right)}$$
$$=3\lim_{x\to +\infty}\frac{\sqrt{1+\frac{1}{9x^5}}}{\frac{1}{x^3}-1}$$
$$=3\cdot \frac{\sqrt{1+0}}{0-1}=\color{red}{-3}$$
A: You can also do
$$\lim_{x \rightarrow -\infty} \frac{\sqrt{9x^6-x}}{x^3+1}$$
$$\sim  \frac{3x^3}{-x^3} = -3$$
This is easier for me to do mentally to check my work; just realize that that the $9x^6$ term grows much faster than the $-x$ term towards postie infinity, so you effectively get $\sqrt{9x^6}$ as you go to infinity, and the addition on the bottom becomes insignificant while the cubic approaches negative infinity. Perhaps a little less rigid the way I word it, but conceptually simple.
