Why do we add a negative sign in front of a square root of a (negative number squared)? Recently while learning calculus in High school, my teacher mentioned how to do limits at infinity  to find horizontal asymptotes by dividing by the largest degree of $x$. For some limits, there is a square root so you have to divide inside the square root, for example if you have the limit:
$\lim_{x \to -\infty} \frac{\sqrt{4x^2+3}}{3x+2}$.
The way my teacher taught it was you divide by $x$, so you get $3 + \frac{2}{x}$ on the bottom. On the top, you are supposed to get $-\sqrt{4 +\frac{3}{x^2}}$ and plug in negative infinity. However, I don't get why you have to add the negative sign in front of the numerator, although I understand it has something to do with the limit being at negative infinity (as this doesn't happen when the limit goes to infinity).
My teacher mentioned something about negative numbers squared and how they work when square rooted, and he mentioned this definition:
$\sqrt{x^2}$ = $-x$ if $x<0$
$\sqrt{x^2}$ = $x$ if $x>0$
However, this doesn't make sense to me. If $x = -5$ for example, wouldn't it be $\sqrt{(-5)^2}$, which is $\sqrt{25}$ which is 5 instead of -5? I'm not sure why the above "definition" or equation exists?
 A: *

*The minus sign in $$-x$$ does not indicate that $-x$ is negative; rather, it operates on $x$ so that the resultant value $-x$ has the opposite sign. (It helps to read $-x$  as “minus $x$” rather than “negative $x$”. )


*Your given definition can be compressed into a single line: $$\sqrt {x^2}=|x|.$$
A: As an equivalent way, instead of dividing by $x$, we can proceed factoring out the leading terms as follows
$$ \frac{\sqrt{4x^2+3}}{3x+2}= \frac{\sqrt{x^2}}{x} \frac{\sqrt{4+\frac 3{x^2}}}{3+\frac 2 x}$$
and since eventually $x<0$ for the leading factor we have
$$\frac{\sqrt{x^2}}{x}= \frac{|x|}{x}=-1$$
as you can easily check plugging in some negative number, as for example
$$\frac{\sqrt{(-5)^2}}{-5}= \frac{5}{-5}=-1$$
As an alternative we can change variable by $y=-x\to \infty$ to obtain
$$\lim_{x\to -\infty}  \frac{\sqrt{4x^2+3}}{3x+2}= \lim_{y\to \infty}  \frac{\sqrt{4y^2+3}}{-3y+2}$$
which leads to the same result.

Edit
For the method proposed “dividing by x” the full steps should be as follow
$$ \frac{\sqrt{4x^2+3}}{3x+2}= \frac{\frac 1 x}{\frac 1 x}  \frac{\sqrt{4x^2+3}}{3x+2} =  \frac{-\sqrt{\frac 1 {x^2}}}{\frac 1 x}  \frac{\sqrt{4x^2+3}}{3x+2} =-\frac{\sqrt{4+\frac 3{x^2}}}{3+\frac 2 x} $$
and the key step is that for $x<0$
$$ \frac 1 x= -\sqrt{\frac 1 {x^2}}\iff \sqrt{x^2}=-x$$
A: If $x=-5$ then yes, $\sqrt{x^2} = 5$.  But we said $x$ is equal to -5, not 5. So it is incorrect to say $\sqrt{x^2} = x$ when $x = -5$. We have $\sqrt{x^2} = -x$ when $x = -5$.
A: I am also in High School, and it is my understanding that if there is a number squared and then square rooted then it is (by definition) just the original number. Taking your example of -5, plugging it in you would get sqrt (-5)^2 = -5. (-5)^2 is 25 and square rooting it is 5 which is not equal to -5. To avoid this, you must specify that if X is negative then sqrt X^2 = -x and if X is positve then sqrt X^2 = X. This would be expressed mathematically using inequalites as seen in your question.
