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?
 A: Divide numerator and denominator by $x$ and recall $x=\sqrt{x^2}$ for $x\ge0$.
A: 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 $n \to \infty$
A: 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.
