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.

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}
