Limit as x approches infinity - Trouble with calculus I have this problem on a practice exam:
$\displaystyle \lim_{x\to\infty} 3x - \sqrt{9x^2+2x+1}$
We are dealing with L'hospitals rule, so when you plug $\infty$ in for $x$ you get 
$\infty - \infty$. I multiplied by the conjugate to get:
$\displaystyle \frac{-2x-1}{3x + \sqrt{9x^2+2x+1}}$
After applying L'hospitals a few times I was lost so I peeked at the solutions sheet and my teacher went from the above step straight to:
$\displaystyle \frac{-2}{3+3} = -\frac{1}{3}$
I'm kind of lost as to how she came to that answer. Can anyone explain to me how that works?

Let $L$ be the given expression, then $L = \dfrac{-2x - 1}{3x + \sqrt{9x^2 + 2x + 1}} = \dfrac{-2 -\dfrac{1}{x}}{3 + \sqrt{9 + \dfrac{2}{x} + \dfrac{1}{x^2}}} \to \dfrac{-2}{6} = \dfrac{-1}{3}$ when $n \to \infty$
