Limit of quotient I recently learned that when you are solving for the limit of a quotient, you have to divide everything by the highest number in the denominator, like this
$$ \lim_{x \to \infty} \frac{\sqrt{4 x^2 - 4}}{x+5} = \lim_{x \to \infty} \frac{\sqrt{4 - \frac{4}{x^2}}}{1 + \frac{5}{x}} = 2.$$
But I don't quite understand how when $x \to - \infty$ , the answer changes to $-2$, since when you divide everything by the highest power, it gives 
$$\frac{\sqrt{4-\frac{4}{x^2}}}{1+\frac{5}{x}},$$ meaning that even if you put negative infinity in the place of $x$, it still only gives $0$ and leaves $2$ as the final answer.
Why does the answer change to $-2$? I understand that it must be $-2$ when I look at the graph, I just don't understand the algebra part of it.  

$\dfrac{\sqrt{4 \cdot 5^2}}{5} 
   = \dfrac{\sqrt{4 \cdot 5^2}}{\sqrt{5^2}}
   = \sqrt{\dfrac{4 \cdot 5^2}{5^2}}
   = 2$
$\dfrac{\sqrt{4 \cdot (-5)^2}}{-5} 
   = \dfrac{\sqrt{4 \cdot (-5)^2}}{-\sqrt{(-5)^2}}
   = -\sqrt{\dfrac{4 \cdot (-5)^2}{(-5)^2}}
   = -2$
If $x < 0$, then $x = -\sqrt{x^2}$.
$\dfrac{\sqrt{4 \cdot x^2}}{x} 
   = \dfrac{\sqrt{4 \cdot x^2}}{-\sqrt{x^2}}
   = -\sqrt{\dfrac{4 \cdot x^2}{x^2}}
   = -2$
