So the question is: $$\lim_{x \to \infty} \frac{\sqrt{x^2 -1}}{2x+1}$$

First of all, I know we have to use Lhopital's rule. However, I just don't know how.

Second of all, I thought in the end we would get just one value. HOWEVER, my teacher started saying that we would get two and therefore the limit wouldn't exist. So, I graphed the function on my calculator and noticed that as x approached infinity, it approached just one value (to the right). There was another graph to the left that approached the same, but negative, value as x approached NEGATIVE infinity.

Well, wait a minute, I said...why are we counting the two values if this problem is asking when x--> infinity (positive infinity, I assumed, since it didn't have a negative sign).

This is what he said: Oh, because in saying x approaches infinity we mean x approaches both directions of infinity (negative and positive). This is because when talking about infinity we don't do "approaching from the left/ approaching from the right of positive or negative infinity...it's just negative or positive infinity" .

I know, it makes no sense. So, that is why, my friends...I am completely confused and have a major headache because I just want to understand and, obviously, I am far from understanding. I thought it's just positive or negative infinity, not 'infinity as a whole'. what does that even mean??

Can someone explain why this limit approaches two different values and therefore doesn't exist, as my teacher so confusingly put it?

  • $\begingroup$ I cannot explain. Your teacher uses the term $\lim_{x\to\infty}$ in a different way than I do. $\endgroup$ Commented Sep 2, 2015 at 3:51
  • $\begingroup$ Your teacher has some 'splaining to do if that is what was said. Second, no, you do not have to use L'Hopital here. Just factor out $x$ from the square root. $\endgroup$
    – zhw.
    Commented Sep 2, 2015 at 3:53
  • $\begingroup$ If anyone has time, could you show the steps leading to an answer? Thank you. And thank you to the above answers. $\endgroup$ Commented Sep 2, 2015 at 3:54
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    $\begingroup$ Your teacher is dead wrong. It is possible to approach positive infinity or negative infinity individually, you do not have to do them both at once. Approaching $+\infty$ means that your x is just getting really big, and approaching $-\infty$ means that your x is getting really big in the negative direction. $\endgroup$
    – Ud779
    Commented Sep 2, 2015 at 4:04
  • $\begingroup$ Thank you all. I'm sad though, I wish I could try to explain to my teacher, but I don't want to be annoying...I've asked him twice and he's given this lengthy explanation twice. $\endgroup$ Commented Sep 2, 2015 at 4:14

2 Answers 2


The idea is to factor out the dominating term:

The limit as $x\to+\infty$: If $x>0$, $$ \frac{\sqrt{x^2-1}}{2x+1}=\frac{x\sqrt{1-1/x^2}}{x(2+1/x)}=\frac{\sqrt{1-1/x^2}}{2+1/x}\to\frac{\sqrt{1}}{2}=\frac{1}{2}, $$ where the limit is taken as $x\to+\infty$.

If you want to study the limit as $x\to-\infty$, you should be a bit more careful, and factor out an absolute value:

If $x<0$, $$ \frac{\sqrt{x^2-1}}{2x+1}=\frac{|x|\sqrt{1-1/x^2}}{x(2+1/x)}=-\frac{\sqrt{1-1/x^2}}{2+1/x}\to-\frac{\sqrt{1}}{2}=-\frac{1}{2}, $$ where the limit is taken as $x\to+\infty$.

  • $\begingroup$ This is the actual answer. My answer shows that L'hopitals is not useful here $\endgroup$
    – Kerry
    Commented Sep 2, 2015 at 4:03
  • $\begingroup$ Thank you! By the way, why do you think my teacher said that it does not exist? because he took "lim x-->infinity" to mean including the value as lim x--> -infinity and the value as lim x---> + infinity. So, he said, because those two values don't equal, then "lim x---> infinity" does not exist. $\endgroup$ Commented Sep 2, 2015 at 4:06
  • $\begingroup$ I think only the teacher can answer that question. I hope everything got clearer. Good luck with the limits! $\endgroup$
    – mickep
    Commented Sep 2, 2015 at 4:08
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    $\begingroup$ Yes, at least I know now my reasoning was right...but it sucks...I'm going to have to put 'DNE' on tests and assignments when I know that's not right. I tried asking him twice, he just explains the same thing. $\endgroup$ Commented Sep 2, 2015 at 4:09

DO NOT use L'Hopitals rule. Here is a demonstration of the general procedure if you tried it:

$$ \lim_{x \to \infty} \dfrac{\sqrt{x^2-1}}{2x+1} = \lim_{x \to \infty} \dfrac{\dfrac{d}{dx}\sqrt{x^2-1}}{\dfrac{d}{dx}2x+1}=\lim_{x \to \infty} \dfrac{x}{2\sqrt{x^2-1}} = \lim_{x \to \infty} \dfrac{\sqrt{x^2-1}}{2x}= \lim_{x \to \infty} \dfrac{\sqrt{x^2(1-\dfrac{1}{x^2})}}{2x}= \lim_{x \to \infty} \dfrac{x\sqrt{1-\dfrac{1}{x^2}}}{2x}= \lim_{x \to \infty} \dfrac{\sqrt{(1-\dfrac{1}{x^2})}}{2} = \dfrac{1}{2}$$

  • $\begingroup$ If you were anyways going to take $x^2$ out of the numerator as a factor, you could have done it as step 1 itself, without ever using the L'Hospital's rule $\endgroup$ Commented Sep 2, 2015 at 4:16
  • $\begingroup$ @DeepakGupta You are right but the question specifically asks how to use L'hopitals rule. " First of all, I know we have to use Lhopital's rule. However, I just don't know how." $\endgroup$
    – Kerry
    Commented Sep 2, 2015 at 4:17
  • $\begingroup$ @Ryan That is precisely why you should not have used it and shown the questioner that the supposition "I know we have to use L'Hospital's rule" is actually incorrect. But that's just my way of looking at it. Cheers. :) $\endgroup$ Commented Sep 2, 2015 at 4:20

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