Calculus 1: Limits at negative infinity of quotients with square root, why introduce a negative when trying to simply

Question

Why do we introduce a negative when trying to simplify? (circled in orange) If x approaches -$\infty$ for both, the negative sign should cancel out..?

Khan acad's working:

Edit, removed my workings as requested & added a follow up: Why don't they do the same here?

Answer 1

It might be helpful to look at the facts/steps all separately:

Square root of a square

For any real number $a$ (negative or not), $a^{2}$ is nonnegative, so we can take the square root of it; $\sqrt{a^{2}}$ makes sense. And the square root function always outputs a nonnegative number.

If $a\ge0$, then $\sqrt{a^{2}}=a$. And if $a<0$, $\sqrt{a^{2}}$ is positive but $a$ is not; we have $\sqrt{a^{2}}=-a$. Combining these two cases together, we have $\boxed{\sqrt{a^{2}}=\left|a\right|}$ for any real $a$. I think that equation is worth memorizing.

Problem 1

In the first problem presented, we have $\sqrt{x^{6}}=\sqrt{\left(x^{3}\right)^{2}}=\left|x^{3}\right|$. Since we're taking a limit as $x\to-\infty$, we only care about negative values of $x$. And the cube of a negative value is negative (e.g. $\left(-2\right)^{3}=-2*4=-8$). When we have the absolute value of a negative number, we negate it to make it positive. So in this context, $\left|x^{3}\right|=-\left(x^{3}\right)$. The order of operations lets us drop the parentheses. So we have $\sqrt{x^{6}}=-x^{3}$ in this context, and can multiply through by $-1$ to get $x^{3}=-\sqrt{x^{6}}$.

Thus, $1=\dfrac{\dfrac{1}{-\sqrt{x^{6}}}}{\dfrac{1}{x^{3}}}$, which helps you solve the limit.

Problem 2

In the second problem presented, we have $\sqrt{x^{4}}=\sqrt{\left(x^{2}\right)^{2}}=\left|x^{2}\right|$. Since we're taking a limit as $x\to-\infty$, we only care about negative values of $x$. But the square of a negative value is positive (e.g. $\left(-2\right)^{2}=4$). When we have the absolute value of a positive number, we just get the original number. So in this context, $\left|x^{2}\right|=x^{2}$. In fact, since the square of a nonnegative number is nonnegative, we would still have $\left|x^{2}\right|=x^{2}$ when $x$ is nonnegative, too. So we have $\sqrt{x^{4}}=x^{2}$ for any real $x$.

Thus, $1=\dfrac{\dfrac{1}{\sqrt{x^{4}}}}{\dfrac{1}{x^{2}}}$, which helps you solve the limit.

Answer 2

If $x<0$, $\sqrt{x^2}=x$ is false and for the same reason $\sqrt{x^6}=x^3$ as well.

Answer 3

I got it! It is because I need to ensure that I'm multiplying by 1. Thanks for all the help along the way. It pushed me to go back to revise exponentials. Two step forward one step back they say!