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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?

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  • $\begingroup$ I cannot explain. Your teacher uses the term $\lim_{x\to\infty}$ in a different way than I do. $\endgroup$ – André Nicolas Sep 2 '15 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. Sep 2 '15 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. $\endgroup$ – helpmeplease Sep 2 '15 at 3:54
  • $\begingroup$ 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 Sep 2 '15 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. $\endgroup$ – helpmeplease Sep 2 '15 at 4:14
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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 $x\to+\infty$.

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  • $\begingroup$ This is the actual answer. My answer shows that L'hopitals is not useful here $\endgroup$ – Kerry Sep 2 '15 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. $\endgroup$ – helpmeplease Sep 2 '15 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 Sep 2 '15 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. $\endgroup$ – helpmeplease Sep 2 '15 at 4:09
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DO NOT use L'Hopitals rule. Here is a demonstration of the general procedure if you tried it:

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

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  • $\begingroup$ The last expression you have looks suspicious. $\endgroup$ – mickep Sep 2 '15 at 3:58
  • $\begingroup$ @mickep fixed it $\endgroup$ – Kerry Sep 2 '15 at 4:00
  • $\begingroup$ There is still an extra $1/2$. $\endgroup$ – mickep Sep 2 '15 at 4:05
  • $\begingroup$ At this point you just factor out an x from the numerator and divide off. $\endgroup$ – Ud779 Sep 2 '15 at 4:06
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    $\begingroup$ @DeepakGupta I agree. I updated my answer with this very clear statement. $\endgroup$ – Kerry Sep 2 '15 at 4:22

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