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.
 A: 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}$?
A: $\lim _{x\rightarrow \infty }x- \sqrt{{x}^{2}-x+2}=\lim _{x\rightarrow \infty }{\frac { ( x-\sqrt {{x}^{2}-x+2})  ( x+\sqrt {{x}^{2}-x+2}) }{ x+\sqrt {{x}^{2}-x+2}}}=\lim _{x\rightarrow \infty }{\frac{x-2}{x+\sqrt {{x}^{2}-x+2}}}=\frac{1}{2}$
A: Multiply by conjugate to get $\operatorname{lim}_{x\to \infty}\frac{x-2}{x+\sqrt{x^2-x+2}}=\operatorname{lim}_{x\to \infty}\frac{1-2/x}{1+\sqrt{1-1/x+2/x^2}}=1/2$
A: Yes rationalize is  a good idea, indeed we obtain
$$x - \sqrt{x^2 - x +2 }=x - \sqrt{x^2 - x +2 }\cdot\frac{x + \sqrt{x^2 - x +2 }}{x + \sqrt{x^2 + x +2 }}=\frac{x^2-x^2+x-2}{x+ \sqrt{x^2 + x +2 }}$$
$$=\frac{x-2}{x+ \sqrt{x^2 + x +2 }}=\frac x x\frac{1-\frac 2x}{1+ \sqrt{1 + \frac1x +\frac2{x^2} }}\to \frac12$$
A: Use Taylor's expansion at order $1$: for $x>0$
\begin{align}
x-\sqrt{x^2-x+2}&=x-x\sqrt{1-\frac1x+\frac2{x^2}}=x-x\sqrt{1-\frac1x+o\Bigl(\frac1{x}\Bigr)}\\
&=x-x\Bigl(1-\frac1{2x}+o\Bigl(\frac1{x}\Bigr)\Bigr)=x-\Bigl(x-\frac12 +o(1)\Bigr) \\
&=\frac12+o(1).
\end{align}
A: As you mentioned calculus, you may also substitute $x=\frac{1}{t}$ and consider the limit for $t\to 0^+$:
\begin{eqnarray*} x - \sqrt{x^2 - x +2 }
& \stackrel{x=\frac{1}{t}}{=} & \frac{1-\sqrt{1-t+2t^2}}{t} \\
& \stackrel{L'Hosp.}{\sim} & -\frac{4t-1}{2\sqrt{1-t+2t^2}} \\
& \stackrel{t\to 0^+}{\rightarrow} & \frac{1}{2} \\
\end{eqnarray*}
A: 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 $$\lim _{x\to -\infty }\left(x- \sqrt{{x}^{2}-x+2}\right)=-\infty.$$
