Let's say we're trying to solve the following limit:
$\lim\limits_{x\to \infty} \sqrt{x^2+x}-x$
One way to do this is to use conjugates which results in the following:
$\lim\limits_{x\to \infty} \frac{\displaystyle x}{\displaystyle\sqrt{x^2+x}+x}$
Here's my problem: For the latter, we can simplify it so we'd have
$\lim\limits_{x\to \infty} \frac{\displaystyle x}{\displaystyle\sqrt{x^2+x} + x} = \lim\limits_{x\to \infty} \frac{\displaystyle x}{\displaystyle x+x} = \lim\limits_{x\to \infty} \frac{\displaystyle x}{\displaystyle 2x} = \frac{\displaystyle 1}{\displaystyle 2}$
but the same thing can't be done to the former. That is, we can't say
$\lim\limits_{x\to \infty} \sqrt{x^2+x}-x = \lim\limits_{x \to \infty} x - x = 0$
I believe this is issue comes from the last equality because $\lim\limits_{x \to \infty} x -x$ is like $\frac {0}{0}$; it's indeterminate but I'm not exactly sure.
EDIT
I'm so sorry I think I didn't phrase my question properly. My bad! I know why the two limits in the title are MUCH different; one tends to infinity whereas the other one has a finite value of $\frac{1}{2}$. I'm confused about why in the limit $\lim\limits_{x \to \infty}\frac{\displaystyle x}{\displaystyle \sqrt{x^2+x} + x}$, we can say the denominator is $2x$ but in the limit $\lim\limits_{x \to \infty} \sqrt{x^2+x} - x$ we can't say the square root is equal to $x$.
Thank you so much in advance!