Limit at negative infinity of a function with a radical $$\lim_{x \to -∞}{\sqrt{x^2 -x} + x}.$$
First I rationalized, to get
$$\frac{x^2-x-x^2}{\sqrt{x^2-x}-x}$$
Then I wanted to factor the whole thing by the largest power(x) in the denominator, to get:
$$\frac{-x/x}{1/x(\sqrt{x^2-x}-x)}$$
After simplifying:
$$\frac{-1}{\sqrt{1-1/x}-1}$$
And evaluating at negative infinity:
$$\frac{-1}{1-0-1}$$
But this is incorrect as the limit would not exist. Where did I go wrong? Am I right in my procedure? 
Thanks
 A: $$\lim _{ x\to -∞ }{ \sqrt { x^{ 2 }-x } +x } =\lim _{ x\to -∞ }{ \frac { \left( \sqrt { x^{ 2 }-x } +x \right) \left( \sqrt { x^{ 2 }-x } -x \right)  }{ \left( \sqrt { x^{ 2 }-x } -x \right)  }  } =\lim _{ x\to -∞ }{ \frac { -x }{ \left( \sqrt { x^{ 2 }-x } -x \right)  }  } =\\ =\lim _{ x\to -∞ }{ \frac { -x }{ \left( \left| x \right| \sqrt { 1-\frac { 1 }{ x }  } -x \right)  }  } =\lim _{ x\to -∞ }{ \frac { -x }{ \left( -x\sqrt { 1-\frac { 1 }{ x }  } -x \right)  }  } =\lim _{ x\to -∞ }{ \frac { -x }{ -x\left( \sqrt { 1-\frac { 1 }{ x }  } +1 \right)  }  } =\\ =\lim _{ x\to -∞ }{ \frac { 1 }{ \left( \sqrt { 1-\frac { 1 }{ x }  } +1 \right)  }  } =\frac { 1 }{ 2 } $$
A: Remember that $\sqrt{A^2}=|A|$.
we have that
$$(\forall x<0)\;\;\; \sqrt{x^2-x}+x=$$
$$|x|\sqrt{1-\frac{1}{x}}+x=$$
$$x(1-\sqrt{1-\frac{1}{x}})=$$
$$\frac{1}{1+\sqrt{1-\frac{1}{x}}}$$
and the limit when $x\to -\infty\;$ is $\frac{1}{2}$.
A: 
Where did I go wrong?

The following is wrong : 
$$\frac{\sqrt{x^2-x}}{x}=\sqrt{1-\frac 1x}$$
Note that for $x\lt 0$
$$\frac{\sqrt{x^2-x}}{x}=\frac{\sqrt{x^2(1-\frac 1x)}}{x}=\frac{\sqrt{x^2}\sqrt{1-\frac 1x}}{x}=\frac{|x|\sqrt{1-\frac 1x}}{x}=\frac{\color{red}{-}x\sqrt{1-\frac 1x}}{x}=\color{red}{-}\sqrt{1-\frac 1x}$$
A: Your limit can be rewritten as
$$
\lim_{x\to +\infty}\sqrt{x^2+x}-x.
$$
Using the fact that, as $x\to \infty$
$$
\sqrt{1+\frac{1}{x}}=1+\frac{1}{2x}+O\left(\frac{1}{x^2}\right)
$$
you conclude that 
$$
\sqrt{x^2+x}-x=x\left(1+\frac{1}{2x}+O\left(\frac{1}{x^2}\right)-1\right) \to \frac{1}{2}.
$$
