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?
Best Answer
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: $$\sqrt {x^2}=|x|.$$