Limit $\lim_{x \to \infty} \frac{\sqrt{x^2 -1}}{2x+1}$ 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?
 A: 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 $x\to+\infty$.
A: DO NOT use L'Hopitals rule. Here is a demonstration of the general procedure if you tried it: 
$$ \lim_{x \to \infty} \dfrac{\sqrt{x^2-1}}{2x+1} = \lim_{x \to \infty} \dfrac{\dfrac{d}{dx}\sqrt{x^2-1}}{\dfrac{d}{dx}2x+1}=\lim_{x \to \infty} \dfrac{x}{2\sqrt{x^2-1}} = \lim_{x \to \infty} \dfrac{\sqrt{x^2-1}}{2x}= \lim_{x \to \infty} \dfrac{\sqrt{x^2(1-\dfrac{1}{x^2})}}{2x}= \lim_{x \to \infty} \dfrac{x\sqrt{1-\dfrac{1}{x^2}}}{2x}= \lim_{x \to \infty} \dfrac{\sqrt{(1-\dfrac{1}{x^2})}}{2} = \dfrac{1}{2}$$
