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?

This is to answer "what is your thought process" (I cannot read LAcarguy's mind so I cannot say what his thought process is).  The trick of multiplying and dividing by the reciprocal of a leading term is a standard trick when we have a limits going to infinity. For instance suppose that we have the following limit, $$\lim_{n\to\infty}\frac{2x^2+3x+5}{5x^2-3x+100}.$$ Well note that the leading term of the numerator is $2x^2$ and the denominator is has leading term $5x^2$. So this suggests that $$\lim_{n\to\infty}\frac{2x^2+3x+5}{5x^2-3x+100}=\frac{2}{5}.$$ Now if we do this properly, we multiply numerator and denominator by $\frac{1}{x^2}$. This gives us, 
$$\lim_{n\to\infty}\frac{2x^2+3x+5}{5x^2-3x+100}=\\
\lim_{n\to\infty}\frac{2x^2+3x+5}{5x^2-3x+100}\frac{1/x^2}{1/x^2}=\\
\lim_{n\to\infty}\frac{2+3/x+5/x^2}{5-3/x+100/x^2}\\ = \frac{2}{5}.$$ Now this is not the problem that you originally posed, so lettuce now look at that.
$$\lim_{n\to\infty}\frac{-2x-1}{3x + \sqrt{9x^2+2x+1}}.$$ The long term behavior of the numerator is the same as the long term behavior of $-2x$, in particular the leading term. Now the denominator is a little more tricky. If we look at the long term behavior of $$\sqrt{9x^2+2x+1}$$, this will tell us what we want to know. But the long term behavior of the polynomial inside the radical is $$9x^2.$$ So then the long term behavior of the radical ,$$\sqrt{9x^2+2x+1}$$ is $$\sqrt{9x^2}=3x$$ as long as $x$ is nonnegative. Therefore the long term behavior of the denominator, $${3x + \sqrt{9x^2+2x+1}}$$ is $$3x+3x=6x.$$ Therefore, the long term behavior of the ratio $$\frac{-2x-1}{3x + \sqrt{9x^2+2x+1}}$$ is $$\frac{-2x}{6x}=-\frac{1}{3}.$$ So we might think that $$\lim_{n\to\infty}\frac{-2x-1}{3x + \sqrt{9x^2+2x+1}}=-\frac{1}{3}.$$ Well now if we do this properly, what we do is multiply the numerator and denominator by $1/x$, and we get the solution that LAcarguy gives.
