Non-rigorous limits to infinity trouble

Question

I had to solve this problem: $$ \lim \limits_{x \to ∞} {x\over \sqrt {3x^2+2}} $$ and I had no idea how to get rid of the square root from the denominator. I googled for some time and found out that you can do this: $$ \lim \limits_{x \to ∞} {x\over \sqrt {3x^2+2}} = \lim \limits_{x \to ∞} {\sqrt {x^2}\over \sqrt {3x^2+2}} = \sqrt {\lim \limits_{x \to ∞} {x^2\over {3x^2+2}}} = ...={1\over \sqrt3} $$ The part that bothers me is $$ x=\sqrt {x^2} $$ because it works like this only if x is not negative. If, for example, x=3, it works, but if x=-3, then it says that x=-3=3. Can we still do this because technically x is non-negative since it's approaching infinity? It bothers me because x really isn't non-negative, we don't know what it is, it just approaches infinity. This is high school calculus so we treat limits very non-rigorously; could someone shed some light on this with something with a bit more rigor?

Answer 1

This is fine as you're looking for the limit as x approaches positive infinity. So you're not interested in any negative values for x. In fact for any given real number N, you don't care what happens when $x \le N$, you're only interested in what happens for $x > N$.

This is rather rigorous, it follows directly from the definition itself
(of what we call limit as x approaches positive infinity).

Answer 2

The part that bothers me is $$ x=\sqrt {x^2} $$ because it works like this only if x is not negative.

A substitution that works for all $\mathbb{R}$ is $$ x = \begin{cases} \sqrt{x^2} & \text{for } x \ge 0 \\ -\sqrt{(-x)^2} & \text{for } x \le 0 \end{cases} $$ Albeit for your specifc limit you will use only the non-negative case at some point, so you do not have to worry about the other case.