what is the limit as $x$ approaches infinity of $x-\sqrt{x^2-x+2}$ $$
\lim_{x \to \infty} \left( x - \sqrt{x^2 - x +2 } \right)
$$

I've tried rationalizing the expression but after repeated applications of L'Hospital's rule, it doesn't feel like I'm getting anywhere.
Note that 

$$ \sqrt{x^2 - x + 2} = \sqrt{x^2 -x + \frac{1}{4} + 1.75} = \sqrt{(x - \frac{1}{2})^2 + 1.75}$$.

What is the limit of $ \sqrt{(x - \frac{1}{2})^2 + 1.75} -  \sqrt{(x - \frac{1}{2})^2}$?
The  limit $\lim\limits_{x \to \infty} \left( x - \sqrt{x^2 - x +2 } \right)$ dones not exist!
Because 
$$\lim _{x\to +\infty }x- \sqrt{{x}^{2}-x+2}
=\lim _{x\to +\infty }{\frac { ( x-\sqrt {{x}^{2}-x+2}) (x+\sqrt {{x}^{2}-x+2}) }{ x+\sqrt {{x}^{2}-x+2}}}$$
$$=\lim_{x\to+\infty }{\frac{x-2}{x+\sqrt {{x}^{2}-x+2}}}=\frac{1}{2}.$$
But $$\lim _{x\to -\infty }\left(x- \sqrt{{x}^{2}-x+2}\right)=-\infty.$$