<s>[INST] 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
 [/INST] $$\lim _{ x\to -∞ }{ \sqrt { x^{ 2 }-x } +x } =\lim _{ x\to -∞ }{ \frac { \left( \sqrt { x^{ 2 }-x } +x \right) \left( \sqrt { x^{ 2 }-x } -x \right)  }{ \left( \sqrt { x^{ 2 }-x } -x \right)  }  } =\lim _{ x\to -∞ }{ \frac { -x }{ \left( \sqrt { x^{ 2 }-x } -x \right)  }  } =\\ =\lim _{ x\to -∞ }{ \frac { -x }{ \left( \left| x \right| \sqrt { 1-\frac { 1 }{ x }  } -x \right)  }  } =\lim _{ x\to -∞ }{ \frac { -x }{ \left( -x\sqrt { 1-\frac { 1 }{ x }  } -x \right)  }  } =\lim _{ x\to -∞ }{ \frac { -x }{ -x\left( \sqrt { 1-\frac { 1 }{ x }  } +1 \right)  }  } =\\ =\lim _{ x\to -∞ }{ \frac { 1 }{ \left( \sqrt { 1-\frac { 1 }{ x }  } +1 \right)  }  } =\frac { 1 }{ 2 } $$
</s>