Why is $\lim\limits_{x\to \infty} \sqrt{x^2+x}-x \ne x - x$ but $\lim\limits_{x\to \infty} \sqrt{x^2+x}+x = x + x$? 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! 
 A: It is meaningless state that $\lim\limits_{x\to \infty} \sqrt{x^2+x}-x \ne x - x$  and $\lim\limits_{x\to \infty} \sqrt{x^2+x}+x = x + x$ we should state that
$$\lim\limits_{x\to \infty} \sqrt{x^2+x}-x = \lim\limits_{x\to \infty} \frac1{2}+o(1/x)=\frac12$$
and
$$\lim\limits_{x\to \infty} \sqrt{x^2+x}+x = \lim\limits_{x\to \infty}  2x+o(1)=\infty$$
the explanation in both case is in binomial first order approximation that is
$$\sqrt{x^2+x}=x\left(1+\frac1x\right)^\frac12= x\left(1+\frac1{2x}+o\left(\frac1x\right)\right)=x+ \frac1{2}+o\left(1\right) $$
which means that for $x$ large we have
$$\sqrt{x^2+x}\sim x+ \frac1{2}$$
and therefore
$$\sqrt{x^2+x}-x \sim \frac 12$$
$$\sqrt{x^2+x}+x \sim 2x+\frac 12$$

Edit
Note that for $\frac{x}{ \sqrt{x^2+x} + x}$ it is not correct to state that the denominator is $2x$ the complete steps are
$$\frac{x}{ \sqrt{x^2+x} + x}=\frac x x \frac{1}{ \sqrt{1+1/x} + 1} \to \frac12$$
For  $\sqrt{x^2+x} - x$ indeed is not correct take$ \sqrt{x^2+x}=x$ what is true is that $\sqrt{x^2+x}\sim x+\frac12$. 
If we use that approximation, it works with both limits. In this particular case first order approximation works and we can use it to evaluate both limits.
A: Hint:
$$ (1\pm f(x))^n \approx 1 \pm nf(x) $$
for $f(x) \to 0$.
The root from your question can be rewritten as:
$$ \sqrt{x^2+x} = x\sqrt{1+\frac{1}{x}} \approx x(1+\frac{1}{2}\cdot\frac{1}{x}) $$
becasue from your limit $ 1/x \to 0 $. From here you can get the answers.
