When do I add a negative sign in front of a square root with the limit approaching negative infinity I am confused on when to add a negative and when not to add a negative when doing limit at negative infinity questions.
Consider this
$$\lim_{x \to -\infty} \frac{\sqrt{4x^4-x}}{2x^2+3}$$
I would solve like this:
$$\lim_{x \to -\infty} \frac{(1/x^2)\sqrt{4x^4-x}}{(1/x^2)({2x^2+3})}$$
$$\lim_{x \to -\infty}  \frac{\sqrt{{4x^4/x^4}-{x/x^4}}}{(1/x^2)({2x^2+3})}$$
Since
$$\sqrt{x^2} = −x \text{ if }  x<0 \\
\sqrt{x^2} = x \text{ if }  x>0$$
Add a negative to the numerator ?? (WRONG)
$$\lim_{x \to -\infty}  \frac{- \sqrt{{4}-{1/x^3}}}{({2+3/x^2})}$$
$$\lim_{x \to -\infty}  \frac{- \sqrt{{4}-0}}{({2+0})}$$
$$\frac{- 2}{({2})}$$
$$=-1 \text{ Wrong answer. Right answer is 1 }$$
The right answer is 1 but I got it wrong because I added negative to the numerator. So can you please tell me when to add the negative sign in front of a square root of x to the power of n?
On the other hand, in this example, the negative seems to be the right thing to do:
$$\lim_{x \to -\infty} \frac{\sqrt{9x^6-x}}{x^3+6}$$
$$\lim_{x \to -\infty} \frac{(1/x^3)\sqrt{9x^6-x}}{(1/x^3)(x^3+6)}$$
$$\lim_{x \to -\infty} \frac{\sqrt{9x^6/x^6-x/x^6}}{(x^3/x^3+6/x^3)}$$
Adding negative to numerator
$$\lim_{x \to -\infty} \frac{-\sqrt{9-1/x^5}}{(1+6/x^3)}$$
$$\frac{-\sqrt{9-0}}{(1+0)}$$
$$-\sqrt{9}$$
$$=-3$$

The key difference is that $x^2$ is even but $x^3$ is odd. Specifically you, in the first example, multiplied by $1/x^2$ and converted this to $\sqrt{1/x^4}$. That’s correct because $x^2$ is always positive. Adding a minus sign is not necessary or correct. For example $(-2)^2=\sqrt{(-2)^4}=\sqrt{16}$ is completely ok since $(-2)^2=4\ge0$. But, for the $x^3$ example, $x^3<0$ when $x<0$ so in fact: $$x^3=-(-x^3)=-\sqrt{(-x^3)^2}=-\sqrt{x^6}$$Is correct.
