Limit of quotient

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.

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: $\lim_{x\to -\infty} \frac{\sqrt{4x^2-4}}{x+5}=-\sqrt{\lim_{x\to-\infty}\frac{4x^2-4}{(x+5)^2}}$

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$ \lim \sqrt{f(x)^2} = \pm \lim f(x)$