Understanding $\sqrt{x^2}$ in a rational expression as x goes to infinity in Calculus

Question

I'm working through a solution for a Calculus problem and am having trouble understanding a couple of points, both highlighted in green:

1) The way that they deduce that $\sqrt{x^2} = x$ doesn't make sense to me. I get that x will be positive as it goes to infinity, but I don't understand why that allows me to violate an algebraic rule that I was taught for years, that $\sqrt{9x^2 + x} \neq \sqrt{9x^2} + \sqrt{x}$

2) How they then use the above to change $x$ into $\sqrt{x^2}$ when they go to divide the numerator and denominator by $x$. I guess if I can accept 1), then I can accept this, but it still seems like slight of hand and not logical math to me.

Can anyone explain why this works? I'm trying to refresh on Calculus after over a decade of not taking (and forgetting virtually all) math, so plain-spoken explanations are a plus. Thank you!

Answer 1

The issue, as you've correctly identified, is the permissible operations we can do with the square root sign. As you say, we can't split it up at the plus sign -- it's not a linear operator like that, so in order to handle it in the formula we need to have a different approach. What we can do with square roots though is multiply and divide them: that does work.

So, we have: \begin{eqnarray} \sqrt{a+b} & \not= & \sqrt{a} + \sqrt{b} \\ \sqrt{ab} & = & \sqrt{a} \cdot \sqrt{b} \\ \sqrt{\frac{a}{b}} & = & \frac{\sqrt{a}}{\sqrt{b}} \end{eqnarray} (It's worth putting some numbers in there to convince yourself of this.)

When we come to divide $\sqrt{9x^2+x}$ by $x$ we need to have $x$ in the form of a square root to be able to use the formulae above. We can do that: $\sqrt{x^2}$ and $x$ have the same value provided $x>0$ and $x\in \mathbb{R}$. The first element you circled in green is stating that this is true and so we can replace $x$ by $\sqrt{x^2}$ without encountering any edge cases or other problems.

This isn't exactly mathematical sleight of hand -- it's exactly the same trick as you use at the start to eliminate the square root from the numerator and move it to the denominator. There you multiplied by $1$, you just picked the specific representation of $1$ that made the rest of the problem much easier to handle. To be absolutely mathematically rigorous you should have noted that the multiplication doesn't work for $x=0$ (since it reduces to $0/0$ which is indeterminate) but that at $x=0$ the original expression is $0$ and continuous.

Answer 2

They do not violate the algebraic rule that you stated. Normally $\sqrt{x^2}=|x|$ but if x is positive $|x|=x$. So therefore $\sqrt{x^2}$ is just $x$. We have not violated any rules here because there is no sum underneath the square root.