Help with limits to infinity

I'm in a beginning calculus class after not touching math of any sort for 4 years. I've forgotten a lot of the more complicated algebra, and I'm struggling with simplifying limits right now. I'm doing my homework right now and coming across a lot of problems I'm getting wrong and I can't understand why, so I am asking for help.

first problem:

1.) $\displaystyle lim x\to \infty \frac {\sqrt{9x^6 - x}}{x^3 + 1}$

To work this one out, I multiplied the top and bottom by $\displaystyle \sqrt{9x^6 - x}$,

so I get

$\displaystyle \frac{9x^6-x}{(x^3+1)(\sqrt{9x^6-x})} $

From there, I divided everything by $\displaystyle x^3$ since that's the highest x value in the denominator in order to get rid of some of the numbers. I wind up with:

$\displaystyle \frac{9x^3-0}{ (1+0)(\sqrt{9-0} } $

Which is $\displaystyle \frac{9x^3}{3 } $ which equals

$\displaystyle 3x^3$

The back of my book however says this answer should just be 3. Why is the x^3 cancelled?

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Here's another I got wrong:

$\displaystyle lim x\to \infty \sqrt {9x^2 + x} - 3x $

All I could think to do was to divide everything by X but I don't even think you're allowed to do that without a denominator (are you?). Anyway when I do that I get

$\displaystyle \sqrt{9+0} - 3 = 0 $

Apparently the answer is supposed to be 1/6, how do you get this?