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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?

2012ssohn
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Travis
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3 Answers3

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Divide numerator and denominator by $x$ and recall $x=\sqrt{x^2}$ for $x\ge0$.

user2850514
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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 $n \to \infty$

DeepSea
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  • I see how this works, but I'm confused as to how you would think to do that. What was your thought process? – Travis Apr 30 '14 at 02:00
  • @Travis: Dividing numerator and dominator by the highest power of $x$ (more generally, dividing by the dominate term) is one of the first techniques ordinarily encountered when beginning a study of limits. In fact, it was a common technique in all the precalculus courses I've taught (such as in finding horizontal asymptotes). Sometimes when you know too much (such as L'Hopital's rule), it's easy to overlook the more basic methods -- see my answer here. – Dave L. Renfro Apr 30 '14 at 15:22
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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.

Baby Dragon
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  • Opps, I didn't see your answer until after I posted my comment above (and scrolled down to see what other answers said). Travis, you can see another example of this method (with all details shown) in my answer to Finding the limit of this sequence using the fastest growing term. – Dave L. Renfro Apr 30 '14 at 15:31
  • @DaveL.Renfro Thank you for that in depth explanation. I've never had someone actually explain that process further than "if the highest powers are the same in both the numerator and denominator than the limit is the leading coefficients divided." Granted I probably zoned out at some point and missed that explanation. I got that question correct on the exam, so all is well. Thanks again. – Travis May 13 '14 at 18:26