<s>[INST] The limit of this function when x goes to minus infinity? I'm looking for the limit of $$\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {1 + x + {x^2}}  - \sqrt {1 - x + {x^2}} } \right)$$
I know it should be -1, but for some reason I always get to 1.
I'm not sure where the difference between $$- \infty$$ and $$\infty$$ is:
$$\sqrt {1 + x + {x^2}}  - \sqrt {1 - x + {x^2}} {{\sqrt {1 + x + {x^2}}  + \sqrt {1 - x + {x^2}} } \over {\sqrt {1 + x + {x^2}}  + \sqrt {1 - x + {x^2}} }} = {{1 + x + {x^2} - \left( {1 - x + {x^2}} \right)} \over {\sqrt {1 + x + {x^2}}  + \sqrt {1 - x + {x^2}} }} = {{2x} \over {\sqrt {1 + x + {x^2}}  + \sqrt {1 - x + {x^2}} }}$$
$$ = {{2x} \over {\sqrt {1 + x + {x^2}}  + \sqrt {1 - x + {x^2}} }}{{{1 \over x}} \over {{1 \over x}}} = {2 \over {\sqrt {{1 \over {{x^2}}} + {1 \over x} + 1}  + \sqrt {{1 \over {{x^2}}} - {1 \over x} + 1} }} = {2 \over {\sqrt {0 + 0 + 1}  + \sqrt {0 - 0 + 1} }} = 1$$
 [/INST] Here is another and perhaps quicker way to proceed.  Use 
$$\sqrt{1+t}=1+\frac12 t +O(t^{-2})$$
Here we have
$$\sqrt{x^2\pm x+1}=\sqrt{x^2(1\pm x^{-1}+x^{-2})}=|x|\left(1\pm \frac1{2x}+O\left(\frac{1}{x^2}\right)\right)$$
whereupon subtracting the "upper" and "lower" signed expressions yields
$$|x|\left(\frac{1}{x}+O\left(\frac{1}{x^2}\right)\right)$$
$$\lim_{x \to -\infty}\left(|x|\left(\frac{1}{x}+O\left(\frac{1}{x^2}\right)\right)\right)=\lim_{x \to -\infty} \frac{|x|}{x}=-1$$
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