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.  
 A: You are making a mistake that is incredibly common, because it is taught very very poorly in school. The "secret" is that 
$$
\sqrt{x^2}=|x|,
$$
and not $x$. When $x\geq0$ you get $x$, but when $x<0$ you get $-x$. 
In your manipulations, you wrote $$ \frac{\sqrt{4 x^2 - 4}}{x+5} =   \frac{\sqrt{4 - \frac{4}{x^2}}}{1 + \frac{5}{x}} .$$ Try that "equality" with $x=-10$, for instance. 
A: As @Martin Argerami said, $\sqrt{x^2} = |x|$, and not $x$. When you want to solve a limit which has square roots, you have to do the following:
$\lim_{x\to \infty} \frac{\sqrt{4x^2-4}}{x+5}=\sqrt{\lim_{x\to\infty}\frac{4x^2-4}{(x+5)^2}}$
The reason we could do this, is that when $x\to \infty$ the denominator is positive, the numerator is positive, hence the whole answer is positive. Our answer is under a square root which means our answer is consistent. (the square root of a real number is non negative.) However, when $x\to -\infty$, you can say that the result is negative, since the numerator is positive and the denominator is negative. In other words: $x + 5 = - \sqrt{(x + 5)^2}$, because $|x + 5| = -(x + 5)$. So:
$\lim_{x\to -\infty} \frac{\sqrt{4x^2-4}}{x+5}=-\sqrt{\lim_{x\to-\infty}\frac{4x^2-4}{(x+5)^2}}$

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